10 infinite series
10 INFINITE SERIES
10.1 Sequences
Preliminary Questions
1. What is a4 for the sequence an = n2 − n?
solution
Substituting n = 4 in the expression for an gives
a4 = 42 − 4 = 12.
2. Which of the following sequences converge to zero?
(a)
n2
n2 + 1
(b) 2n
(c)
−1 n
2
solution
(a) This sequence does not converge to zero:
lim
n2
n→∞ n2 + 1
= lim
x2
x→∞ x 2 + 1
= lim
1
x→∞ 1 + 1
x2
=
1
= 1.
1+0
(b) This sequence does not converge to zero: this is a geometric sequence with r = 2 > 1; hence, the sequence diverges
to ∞.
(c) Recall that if |an | converges to 0, then an must also converge to zero. Here,
n n
−1 = 1 ,
2 2
which is a geometric sequence with 0 < r < 1; hence, ( 12 )n converges to zero. It therefore follows that (− 12 )n converges
to zero.
√
3. Let an be the nth decimal approximation to 2. That is, a1 = 1, a2 = 1.4, a3 = 1.41, etc. What is lim an ?
n→∞
solution
√
lim an = 2.
n→∞
4. Which of the following sequences is defined recursively?
√
(a) an = 4 + n
(b) bn = 4 + bn−1
solution
(a) an can be computed directly, since it depends on n only and not on preceding terms. Therefore an is defined explicitly
and not recursively.
(b) bn is computed in terms of the preceding term bn−1 , hence the sequence {bn } is defined recursively.
5. Theorem 5 says that every convergent sequence is bounded. Determine if the following statements are true or false
and if false, give a counterexample.
(a) If {an } is bounded, then it converges.
(b) If {an } is not bounded, then it diverges.
(c) If {an } diverges, then it is not bounded.
solution
(a) This statement is false. The sequence an = cos πn is bounded since −1 ≤ cos πn ≤ 1 for all n, but it does not
converge: since an = cos nπ = (−1)n , the terms assume the two values 1 and −1 alternately, hence they do not approach
one value.
(b) By Theorem 5, a converging sequence must be bounded. Therefore, if a sequence is not bounded, it certainly does
not converge.
(c) The statement is false. The sequence an = (−1)n is bounded, but it does not approach one limit.
633
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C H A P T E R 10
INFINITE SERIES
Exercises
1. Match each sequence with its general term:
a1 , a2 , a3 , a4 , . . .
General term
(a) 12 , 23 , 34 , 45 , . . .
(i) cos πn
(b) −1, 1, −1, 1, . . .
n!
(ii) n
2
(c) 1, −1, 1, −1, . . .
(iii) (−1)n+1
(d) 21 , 24 , 68 , 24
16 . . .
(iv)
n
n+1
solution
(a) The numerator of each term is the same as the index of the term, and the denominator is one more than the numerator;
n , n = 1, 2, 3, . . . .
hence an = n+1
(b) The terms of this sequence are alternating between −1 and 1 so that the positive terms are in the even places. Since
cos πn = 1 for even n and cos πn = −1 for odd n, we have an = cos πn, n = 1, 2, . . . .
(c) The terms an are 1 for odd n and −1 for even n. Hence, an = (−1)n+1 , n = 1, 2, . . .
(d) The numerator of each term is n!, and the denominator is 2n ; hence, an = 2n!n , n = 1, 2, 3, . . . .
In Exercises 3–12, calculate
the first four terms of the sequence, starting with n = 1.
1
Let ann =
for n = 1, 2, 3, . . . . Write out the first three terms of the following sequences.
2n − 1
3
3. c(a)
n =b = a
(b) cn = an+3
nn! n+1
solution
n = 1, 2, 3, 4 in the formula for cn gives (d) en = 2an − an+1
(c) dn =Setting
an2
c1 =
3
31
= = 3,
1!
1
c3 =
27
9
33
=
= ,
3!
6
2
c2 =
9
32
= ,
2!
2
c4 =
81
27
34
=
=
.
4!
24
8
5. a1 = 2, an+1 = 2an2 − 3
(2n − 1)!
bn = For n = 1, 2, 3 we have:
solution
n!
a2 = a1+1 = 2a12 − 3 = 2 · 4 − 3 = 5;
a3 = a2+1 = 2a22 − 3 = 2 · 25 − 3 = 47;
a4 = a3+1 = 2a32 − 3 = 2 · 2209 − 3 = 4415.
The first four terms of {an } are 2, 5, 47, 4415.
7. bn = 5 + cos πn
1
1, nb=
b1 = For
n=
solution
1, b2,n−1
3, 4+we have
bn−1
b1 = 5 + cos π = 4;
b2 = 5 + cos 2π = 6;
b3 = 5 + cos 3π = 4;
b4 = 5 + cos 4π = 6.
The first four terms of {bn } are 4, 6, 4, 6.
1 1
1
9. cn =
+2n+1
+ ··· +
cn 1=+(−1)
2 3
n
solution
c1 = 1;
3
1
= ;
2
2
3 1
11
1 1
;
=1+ + = + =
2 3
2 3
6
1 1 1
11 1
25
=1+ + + =
+ =
.
2 3 4
6
4
12
c2 = 1 +
c3
c4
May 23, 2011
S E C T I O N 10.1
Sequences
635
11. b1 = 2, b2 = 3, bn = 2bn−1 + bn−2
an = n + (n + 1) + (n + 2) + · · · + (2n)
solution We need to find b3 and b4 . Setting n = 3 and n = 4 and using the given values for b1 and b2 we obtain:
b3 = 2b3−1 + b3−2 = 2b2 + b1 = 2 · 3 + 2 = 8;
b4 = 2b4−1 + b4−2 = 2b3 + b2 = 2 · 8 + 3 = 19.
The first four terms of the sequence {bn } are 2, 3, 8, 19.
13. Find a formula for the nth term of each sequence.
cn = n-place decimal approximation to e
1 −1
1
2 3 4
(a) ,
,
, ...
(b) , , , . . .
1 8 27
6 7 8
solution
(a) The denominators are the third powers of the positive integers starting with n = 1. Also, the sign of the terms is
alternating with the sign of the first term being positive. Thus,
1
(−1)1+1
a1 = 3 =
;
1
13
1
(−1)2+1
a2 = − 3 =
;
2
23
1
(−1)3+1
a3 = 3 =
.
3
33
This rule leads to the following formula for the nth term:
an =
(−1)n+1
.
n3
(b) Assuming a starting index of n = 1, we see that each numerator is one more than the index and the denominator is
four more than the numerator. Thus, the general term an is
an =
n+1
.
n+5
In Exercises
15–26,
Theorem
to determine
the7.limit
of the sequence or state that the sequence diverges.
Suppose
thatuse
lim
an = 41and
lim bn =
Determine:
n→∞
n→∞
12 (an + bn )
15. a(a)
n = lim
(b) lim an3
n→∞
n→∞
solution We have an = f (n) where f (x) = 12; thus,
(c) lim cos(πbn )
(d) lim (an2 − 2an bn )
n→∞
n→∞
lim an = lim f (x) = lim 12 = 12.
n→∞
x→∞
x→∞
5n − 1
17. bn =
4
= 20+−9 2
an 12n
n
5x − 1
; thus,
solution We have bn = f (n) where f (x) =
12x + 9
lim
5n − 1
n→∞ 12n + 9
= lim
5x − 1
=
x→∞ 12x + 9
5
.
12
19. cn = −2−n
4 + n − 3n2
an = We have
solution
c = f (n) where f (x) = −2−x ; thus,
4n2 + 1n
1
lim −2−n = lim −2−x = lim − x = 0.
n→∞
x→∞
x→∞ 2
21. cn = 9n n
1
zn = We have cn = f (n) where f (x) = 9x ; thus,
solution
3
lim 9n = lim 9x = ∞
n→∞
x→∞
Thus, the sequence 9n diverges.
n
23. an = −1/n
zn =n10
2+1
x
; thus,
solution We have an = f (n) where f (x) = 2
x +1
lim n→∞
x
1
x
1
1
= lim = lim √ x
= lim = lim = √
= 1.
2
2
x→∞
x→∞
x→∞
x→∞
x +1
1+0
x +1
n2 + 1
x2 + 1
1 + 12
2
n
x
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x
x
636
C H A P T E R 10
INFINITE SERIES
12nn + 2
25. an =
an ln
= −93+ 4n
n +1
12x + 2
solution We have an = f (n) where f (x) = ln
; thus,
−9 + 4x
12n + 2
12x + 2
12x + 2
= lim ln
= ln lim
= ln 3
lim ln
n→∞
x→∞
x→∞ −9 + 4x
−9 + 4n
−9 + 4x
In Exercises 27–30, use Theorem
4 to determine the limit of the sequence.
rn = ln n − ln(n2 + 1)
1
27. an = 4 +
n
solution We have
lim 4 +
n→∞
Since
1
1
= lim 4 + = 4
n x→∞
x
√
x is a continuous function for x > 0, Theorem 4 tells us that
1
1 √
lim 4 + =
lim 4 + = 4 = 2
n→∞
n→∞
n
n
n3
= −1
e4n/(3n+9)
an cos
29. an =
2n3 + 1
solution We have
n3
lim
n→∞ 2n3 + 1
=
1
2
Since cos−1 (x) is continuous for all x, Theorem 4 tells us that
π
n3
n3
−1
−1
lim
lim cos
= cos
= cos−1 (1/2) =
3
3
n→∞
n→∞
3
2n + 1
2n + 1
n
31. Let an = −1 . −n
Find a number M such that:
an = tan
n + 1(e )
(a) |an − 1| ≤ 0.001 for n ≥ M.
(b) |an − 1| ≤ 0.00001 for n ≥ M.
Then use the limit definition to prove that lim an = 1.
n→∞
solution
(a) We have
n − (n + 1) −1 n
= 1 .
− 1 = =
|an − 1| = n+1
n + 1 n + 1 n + 1
1 ≤ 0.001, that is, n ≥ 999. It follows that we can take M = 999.
Therefore |an − 1| ≤ 0.001 provided n+1
1 ≤ 0.00001, that is, n ≥ 99999. It follows that we can take M = 99999.
(b) By part (a), |an − 1| ≤ 0.00001 provided n+1
We now prove formally that lim an = 1. Using part (a), we know that
n→∞
|an − 1| =
1
< ,
n+1
provided n > 1 − 1. Thus, Let > 0 and take M = 1 − 1. Then, for n > M, we have
|an − 1| =
1
1
<
= .
n+1
M +1
n
33. Use the limit definition
to prove that lim n−2 = 0.
n→∞
Let bn = 13 .
solution
see that
(a) FindWe
a value
of M such that |bn | ≤ 10−5 for n ≥ M.
(b) Use the limit definition to prove that lim bn = 0. 1 1
−2 − 0| = n→∞
|n
n2 = n2 < May 23, 2011
S E C T I O N 10.1
Sequences
637
provided
1
n> √ .
Thus, let > 0 and take M = √1 . Then, for n > M, we have
1
1
1
|n−2 − 0| = 2 = 2 < 2 = .
n
n
M
In Exercises 35–62, use the appropriate limit laws andn theorems to determine the limit of the sequence or show that it
= 1.
Use the limit definition to prove that lim
diverges.
n→∞ n + n−1
1 n
35. an = 10 + −
9
solution
By the Limit Laws for Sequences we have:
1 n
1 n
1 n
lim 10 + −
= 10 + lim −
.
= lim 10 + lim −
n→∞
n→∞
n→∞
n→∞
9
9
9
Now,
n
n 1 n
1
1
≤ −
≤
.
9
9
9
−
Because
n
1
= 0,
n→∞ 9
lim
by the Limit Laws for Sequences,
n
n
1
1
= − lim
= 0.
n→∞
n→∞
9
9
lim −
Thus, we have
1 n
= 0,
lim −
n→∞
9
and
1 n
10 + −
= 10 + 0 = 10.
n→∞
9
lim
37. cn = 1.01√n
√
dn = n + 3 − n
solution Since cn = f (n) where f (x) = 1.01x , we have
lim 1.01n = lim 1.01x = ∞
n→∞
x→∞
so that the sequence diverges.
39. an = 21/n
2
bn = e1−n
solution
Because 2x is a continuous function,
lim 21/n = lim 21/x = 2limx→∞ (1/x) = 20 = 1.
n→∞
x→∞
9n
41. cn =
= n1/n
bn n!
solution
For n ≥ 9, write
cn =
9 9
9 9 9
9
9
9n
= · ··· ·
·
···
·
n!
1
2
9
10
11
n
−
1
n
call this C
Each factor is less than 1
Then clearly
0≤
May 23, 2011
9n
9
≤C
n!
n
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C H A P T E R 10
INFINITE SERIES
since each factor after the first nine is < 1. The squeeze theorem tells us that
9
9n
9
≤ lim C = C lim
=C·0=0
n→∞ n!
n→∞ n
n→∞ n
lim 0 ≤ lim
n→∞
so that limn→∞ cn = 0 as well.
3n2 +
n+2
43. an =
82n
an =2n2 − 3
n!
solution
3n2 + n + 2
3x 2 + x + 2
3
= lim
= .
2
n→∞ 2n − 3
x→∞ 2x 2 − 3
2
lim
cos n √
45. an =
n
an =n √
n
+
4 ≤ cos n ≤ 1 the following holds:
solution Since −1
−
cos n
1
1
≤
≤ .
n
n
n
We now apply the Squeeze Theorem for Sequences and the limits
lim −
n→∞
1
1
= lim
=0
n n→∞ n
to conclude that lim cosn n = 0.
n→∞
n
47. dn = ln 5 − ln
n!
(−1)n
cn = Note
√ that
solution
n
dn = ln
5n
n!
so that
edn =
5n
n!
so
lim edn = lim
n→∞
5n
n→∞ n!
=0
by the method of Exercise 41. If dn converged, we could, since f (x) = ex is continuous, then write
lim edn = elimn→∞ dn = 0
n→∞
which is impossible. Thus {dn } diverges.
4 1/3
dn =2 ln(n
+ 22 + 4) − ln(n2 − 1)
49. an =
n
1/3
solution Let an = 2 + 42
. Taking the natural logarithm of both sides of this expression yields
n
4
4 1/3
1
= ln 2 + 2 .
ln an = ln 2 + 2
3
n
n
Thus,
4 1/3
1
4
1
4
1
ln 2 + 2
lim ln 2 + 2 = ln lim 2 + 2
=
n→∞ 3
x→∞
3 x→∞
3
n
x
x
lim ln an = lim
n→∞
=
1
1
ln (2 + 0) = ln 2 = ln 21/3 .
3
3
Because f (x) = ex is a continuous function, it follows that
1/3
lim an = lim eln an = elimn→∞ (ln an ) = eln 2
n→∞
n→∞
= 21/3 .
2n + 1
51. cn = ln
2
3n−1
+ 41 −
bn = tan
n
solution Because f (x) = ln x is a continuous function, it follows that
2
2x + 1
2x + 1
= ln lim
= ln .
lim cn = lim ln
n→∞
x→∞
x→∞ 3x + 4
3x + 4
3
May 23, 2011
S E C T I O N 10.1
Sequences
639
en
53. yn =
cn 2=n
n
nn + n1/nn
n
solution 2e n = 2e and 2e > 1. By the Limit of Geometric Sequences,we conclude that limn→∞ 2e = ∞. Thus,
the given sequence diverges.
en +n (−3)n
55. yn =
an = n5n
2
solution
e n
−3 n
en + (−3)n
= lim
+ lim
lim
n→∞
n→∞ 5
n→∞
5n
5
assuming both limits on the right-hand side exist. But by the Limit of Geometric Sequences, since
−1 <
−3
e
<0< <1
5
5
both limits on the right-hand side are 0, so that yn converges to 0.
π
57. an = n sin
n 3
−n
(−1)
n n +2
bn =
+ 4−n on Sequences Defined by a Function, we have
solution By 3n
the3 Theorem
lim n sin
n→∞
Now,
π
π
= lim x sin .
x→∞
n
x
cos πx − π2
sin πx
π
π
x
lim x sin = lim
=
lim
=
lim
π
cos
x→∞
x→∞ 1
x→∞
x→∞
x
x
− 12
x
x
π
= π lim cos = π cos 0 = π · 1 = π.
x→∞
x
Thus,
lim n sin
n→∞
π
= π.
n
3 − 4n
n!
59. bn =
bn 2=+ 7n· 4n
π
solution Divide the numerator and denominator by 4n to obtain
n
4
3
3
3 − 4n
4n − 4n = 4n − 1 .
an =
=
n
2
7·4
2
2 + 7 · 4n
4n + 4n
4n + 7
Thus,
3 −1
limx→∞ 43x − 1
3 limx→∞
x
4
=
lim an = lim 2
=
n→∞
x→∞
2
2 limx→∞
limx→∞ 4x + 7
4x + 7
1
4x − limx→∞ 1 = 3 · 0 − 1 = − 1 .
1
2·0+7
7
4x − limx→∞ 7
1 n
61. an = 1 +3 − 4n
an = n
2 + 7 · 3n
solution Taking the natural logarithm of both sides of this expression yields
n
ln 1 + 1
n
1
1
ln an = ln 1 +
=
= n ln 1 +
.
1
n
n
n
Thus,
lim (ln an ) = lim
n→∞
ln 1 + x1
x→∞
1
x
= lim
x→∞
d
dx
1 · − 1
1
2
ln 1 + x1
x
1+ x
1
1
= lim
= 1.
= lim
=
1
1
x→∞
x→∞
d
1
1
+
0
1+ x
− 2
dx
x
x
Because f (x) = ex is a continuous function, it follows that
lim an = lim eln an = elimn→∞ (ln an ) = e1 = e.
n→∞
1 n
an = 1 + 2
n
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n→∞
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C H A P T E R 10
INFINITE SERIES
In Exercises 63–66, find the limit of the sequence using L’Hôpital’s Rule.
(ln n)2
n
solution
63. an =
d (ln x)2
2 ln x
(ln n)2
(ln x)2
2 ln x
x
= lim
= lim
= lim dx d
= lim
n→∞
x→∞
x→∞
x→∞
x→∞
n
x
1
x
dx x
lim
d 2 ln x
2
2
= lim x = lim
= lim dx d
=0
x→∞
x→∞
x→∞
1
x
dx x
65. cn = n √n2 + 1 − n1
bn = n ln 1 +
solution
n
x
x2 + 1 − x
x2 + 1 + x
lim n
n2 + 1 − n = lim x
x 2 + 1 − x = lim
n→∞
x→∞
x→∞
x2 + 1 + x
= lim x→∞
= lim
x→∞
x
x2 + 1 + x
1+
1
x2
x 2 +1
d x
1
= lim
dx
x→∞ d x 2 + 1 + x
x→∞ 1 + √ x
dx
x 2 +1
= lim
= lim
x→∞
1+
1
1
1+(1/x 2 )
=
1
2
In Exercises 67–70,
Theorem to evaluate lim an by verifying the given inequality.
3 use the Squeeze
n→∞
dn = n2 n3 + 1 − n
1
1
1
67. an = , √
≤ an ≤ √
4
2
4
8
2n
2n
n +n
4
8
solution For all n > 1 we have n < n , so the quotient √ 41 8 is smaller than √ 41 4 and larger than √ 81
n +n
That is,
n +n
n +n8
.
1
1
1
an < = √
; and
= √
2n2
n4 · 2
n4 + n4
1
1
1
= √
= √
.
an > 4
8
8
8
2n
2n
n +n
Now, since lim √
n→∞
1
2n4
1
= lim √
= 0, the Squeeze Theorem for Sequences implies that lim an = 0.
n→∞ 2n2
n→∞
1/n · 3
69. an = (2n + 31n )1/n , 3 ≤1an ≤ (2 · 3n )1/n =
1 2
n
n
n
c
=
+
+
·
·
·
+
,
n
solution Clearly
2 + 3 2≥ 3 for all n ≥ 1.2Therefore:
n2 + 1
n +2
n +n
n
n
(2n + 3n )1/n ≥ (3n )1/n = 3.
≤ cn ≤ 2
2
n +n
n +1
Also 2n + 3n ≤ 3n + 3n = 2 · 3n , so
(2n + 3n )1/n ≤ (2 · 3n )1/n = 21/n · 3.
Thus,
3 ≤ (2n + 3n )1/n ≤ 21/n · 3.
Because
lim 21/n · 3 = 3 lim 21/n = 3 · 1 = 3
n→∞
n→∞
and limn→∞ 3 = 3, the Squeeze Theorem for Sequences guarantees
lim (2n + 3n )1/n = 3.
n→∞
ofn the
isn equivalent
to the assertion lim an = L? Explain.
1/n
(n + 10
)1/nfollowing
, 10 ≤ astatements
an =Which
n ≤ (2 · 10 )
n→∞
(a) For every > 0, the interval (L − , L + ) contains at least one element of the sequence {an }.
(b) For every > 0, the interval (L − , L + ) contains all but at most finitely many elements of the sequence {an }.
solution Statement (b) is equivalent to Definition 1 of the limit, since the assertion “|an − L| < for all n > M”
means that L − < an < L + for all n > M; that is, the interval (L − , L + ) contains all the elements an except
(maybe) the finite number of elements a1 , a2 , . . . , aM .
71.
May 23, 2011
S E C T I O N 10.1
Sequences
641
Statement (a) is not equivalent to the assertion lim an = L. We show this, by considering the following sequence:
n→∞
⎧
1
⎪
⎪
⎨
n
an =
⎪
⎪
⎩1 + 1
n
for odd n
for even n
Clearly for every > 0, the interval (−, ) = (L − , L + ) for L = 0 contains at least one element of {an }, but the
sequence diverges (rather than converges to L = 0). Since the terms in the odd places converge to 0 and the terms in the
even places converge to 1. Hence, an does not approach one limit.
3n2 1
73. Show
thatthat
an =
Find an upper bound.
is decreasing.
Show
an n=2 + 2 is increasing.
2n + 1
solution
2
. Then
Let f (x) = 3x
2
x +2
f (x) =
6x(x 2 + 2) − 3x 2 · 2x
2
(x 2 + 2)
=
12x
2
(x 2 + 2)
.
f (x) > 0 for x > 0, hence f is increasing on this interval. It follows that an = f (n) is also increasing. We now show
that M = 3 is an upper bound for an , by writing:
3n2
3n2 + 6
3(n2 + 2)
≤ 2
=
= 3.
an = 2
n +2
n +2
n2 + 2
That is, an ≤ 3 for all n.
75. Give an example of √
a3 divergent sequence {an } such that lim |an | converges.
n→∞
Show that an = n + 1 − n is decreasing.
solution Let an = (−1)n . The sequence {an } diverges because the terms alternate between +1 and −1; however, the
sequence {|an |} converges because it is a constant sequence, all of whose terms are equal to 1.
77. Using the limit definition, prove that if {an } converges and {bn } diverges, then {an + bn } diverges.
Give an example of divergent sequences {an } and {bn } such that {an + bn } converges.
solution We will prove this result by contradiction. Suppose limn→∞ an = L1 and that {an + bn } converges to a
limit L2 . Now, let > 0. Because {an } converges to L1 and {an + bn } converges to L2 , it follows that there exist numbers
M1 and M2 such that:
2
| (an + bn ) − L2 | <
2
|an − L1 | <
for all n > M1 ,
for all n > M2 .
Thus, for n > M = max{M1 , M2 },
|an − L1 | <
2
and | (an + bn ) − L2 | <
.
2
By the triangle inequality,
|bn − (L2 − L1 )| = |an + bn − an − (L2 − L1 )| = |(−an + L1 ) + (an + bn − L2 )|
≤ |L1 − an | + |an + bn − L2 |.
Thus, for n > M,
|bn − (L2 − L1 ) | <
+ = ;
2 2
that is, {bn } converges to L2 − L1 , in contradiction to the given data. Thus, {an + bn } must diverge.
79. Theorem 1 states that if lim f (x) = L, then the sequence an = f (n) converges and lim an = L. Show that the
x→∞
n→∞
limit L, then there exists a
Use the limit definition
to prove that if {an } is a convergent sequence of integers with
converse
is false.
In other
find a function f (x) such that an = f (n) converges but lim f (x) does not exist.
number
M such
that awords,
n = L for all n ≥ M.
x→∞
solution Let f (x) = sin πx and an = sin πn. Then an = f (n). Since sin πx is oscillating between −1 and 1 the
limit lim f (x) does not exist. However, the sequence {an } is the constant sequence in which an = sin π n = 0 for all n,
x→∞
hence it converges to zero.
Use the limit definition to prove that the limit does not change if a finite number of terms are added or removed
from a convergent sequence.
May 23, 2011
642
C H A P T E R 10
INFINITE SERIES
81. Let bn = an+1 . Use the limit definition to prove that if {an } converges, then {bn } also converges and lim an =
n→∞
lim bn .
n→∞
solution Suppose {an } converges to L. Let bn = an+1 , and let > 0. Because {an } converges to L, there exists an
M such that |an − L| < for n > M . Now, let M = M − 1. Then, whenever n > M, n + 1 > M + 1 = M . Thus,
for n > M,
|bn − L| = |an+1 − L| < .
Hence, {bn } converges to L.
√ that lim a exists if and only if there
nonzero. Show
Let {an } be a sequence such that lim |an | exists√and is √
n
increasing
and bounded above by
83. Proceed as in Example 12 to show that
the sequence 3, 3 3, 3 3 3, . . . isn→∞
n→∞
does
not
change
for
n
>
M.
exists
an
integer
M
such
that
the
sign
of
a
n its value.
M = 3. Then prove that the limit exists and find
solution This sequence is defined recursively by the formula:
an+1 =
3an ,
a1 =
√
3.
Consider the following inequalities:
√
√
√
3a1 = 3 3 > 3 = a1
√
√
a3 = 3a2 > 3a1 = a2
√
√
a4 = 3a3 > 3a2 = a3
a2 =
⇒
a2 > a1 ;
⇒
a3 > a2 ;
⇒
a4 > a3 .
In general, if we assume that ak > ak−1 , then
ak+1 =
3ak > 3ak−1 = ak .
Hence, by mathematical
{an } is increasing.
√
√ induction, an+1 > an for all n; that is, the sequence
Because an+1 = 3an , it follows that an ≥ 0 for all n. Now, a1 = 3 < 3. If ak ≤ 3, then
√
ak+1 = 3ak ≤ 3 · 3 = 3.
Thus, by mathematical induction, an ≤ 3 for all n.
Since {an } is increasing and bounded, it follows by the Theorem on Bounded Monotonic Sequences that this sequence
is converging. Denote the limit by L = limn→∞ an . Using Exercise 81, it follows that
√
L = lim an+1 = lim 3an = 3 lim an = 3L.
n→∞
n→∞
n→∞
√
Thus, L2 = 3L, so L =√0 or L = 3. Because the sequence is increasing, we have an ≥ a1 = 3 for all n. Hence, the
limit also satisfies L ≥ 3. We conclude that the appropriate solution is L = 3; that is, lim an = 3.
n→∞
Let {an } be the sequence defined recursively by
Further
Insights and Challenges
√
n
n/2 by observing that half of the factors of n! are greater
n! = ∞. Hint: Verify that
n!0,≥ (n/2)
a0 =
an+1
= 2 + an
n→∞
than or equal to n/2.
√
√
n √
n/2
Thus, a1 We
= show
2, that
a2 =n! ≥2 + 2,. For
a3 n=≥ 42even,
+ 2we
+ have:
2, . . . .
solution
2
(a) Show that if an < 2, then an+1 < 2. Conclude
by
induction
that
2 for all n.
n an <
n n
+ 1 by· ·induction
· · · n ≥ that+{a1n } ·is· ·increasing.
· · n.
·
n! a=n 1≤· a· n+1
· · · . Conclude
(b) Show that if an < 2, then
√
L= 2lim 2a exists.
2
Then
compute
by showing
(c) Use (a) and (b) to conclude that
L
that L = 2 + L.
nn
n
n
85. Show that lim
2
factorsn→∞
2
factors
2
factors
Since each one of the n2 factors is greater than n2 , we have:
n! ≥
n
n n n/2
n
+ 1 · ··· · n ≥ · ··· · =
.
2
2
2 2
n
2
n
2
factors
factors
For n ≥ 3 odd, we have:
n+1
n−1 n+1
· ··· · n ≥
· · · · · n.
·
n! = 1 · · · · ·
2
2
2 n−1
2
May 23, 2011
factors
n+1
2
factors
n+1
2
factors
S E C T I O N 10.1
Sequences
643
n
Since each one of the n+1
2 factors is greater than 2 , we have:
n! ≥
n n (n+1)/2 n n/2 n n n/2
n
n+1
· ··· · n ≥ · ··· · =
≥
=
.
2
2
2
2
2 2 2
n+1
2
factors
n+1
2
factors
n/2
. Thus,
In either case we have n! ≥ n2
√
n
n! ≥
Since lim
n→∞
n = ∞, it follows that lim
2
n→∞
n
.
2
√
√
n
n! = ∞. Thus, the sequence an = n n! diverges.
87. Given positive√
numbers a1 < b1 , define two sequences recursively by
n
n!
.
Let bn =
an + bn
n
an+1 = an bn ,
bn+1 =
n
2
1 k
=
ln
.
(a)
Show
that
ln
b
n
(a) Show that an ≤ bn for all
n n (Figure
n 13).
k=1and {b } is decreasing.
(b) Show that {an } is increasing
n
bn − an 1
converges
to
(b)
Show
that
ln
b
(c) Show that bn+1 − nan+1 ≤
. ln x dx, and conclude that bn → e−1 .
2 0
(d) Prove that both {an } and {bn } converge and have the same limit. This limit, denoted AGM(a1 , b1 ), is called the
arithmetic-geometric √
mean of a1 and b1 .
(e) Estimate AGM(1, 2) to three decimal places.
Geometric Arithmetic
mean
mean
an
a n+1
b n+1
AGM(a 1, b 1)
bn
x
FIGURE 13
solution
(a) Examine the following:
√ 2
√ 2
√
√ √
an − 2 an bn + bn
an + bn an + bn − 2 an bn
bn+1 − an+1 =
− an bn =
=
2
2
2
√
√ 2
an − bn
≥ 0.
=
2
We conclude that bn+1 ≥ an+1 for all n > 1. By the given information b1 > a1 ; hence, bn ≥ an for all n.
(b) By part (a), bn ≥ an for all n, so
√
an+1 = an bn ≥ an · an = an2 = an
for all n. Hence, the sequence {an } is increasing. Moreover, since an ≤ bn for all n,
bn+1 =
bn + bn
2bn
an + bn
≤
=
= bn
2
2
2
for all n; that is, the sequence {bn } is decreasing.
(c) Since {an } is increasing, an+1 ≥ an . Thus,
bn+1 − an+1 ≤ bn+1 − an =
an + bn
an + bn − 2an
bn − an
− an =
=
.
2
2
2
Now, by part (a), an ≤ bn for all n. By part (b), {bn } is decreasing. Hence bn ≤ b1 for all n. Combining the two inequalities
we conclude that an ≤ b1 for all n. That is, the sequence {an } is increasing and bounded (0 ≤ an ≤ b1 ). By the Theorem
on Bounded Monotonic Sequences we conclude that {an } converges. Similarly, since {an } is increasing, an ≥ a1 for all
n. We combine this inequality with bn ≥ an to conclude that bn ≥ a1 for all n. Thus, {bn } is decreasing and bounded
(a1 ≤ bn ≤ b1 ); hence this sequence converges.
To show that {an } and {bn } converge to the same limit, note that
− an−1
−a
b
b
b −a
bn − an ≤ n−1
≤ n−2 2 n−2 ≤ · · · ≤ 1 n−1 1 .
2
2
2
Thus,
lim (bn − an ) = (b1 − a1 ) lim
n→∞
May 23, 2011
1
n→∞ 2n−1
= 0.
644
C H A P T E R 10
INFINITE SERIES
(d) We have
an+1 =
an bn ,
a1 = 1;
bn+1 =
an + bn
,
2
b1 =
√
2
Computing the values of an and bn until the first three decimal digits are equal in successive terms, we obtain:
√
a2 = a1 b1 = 1 · 2 = 1.1892
√
1+ 2
a1 + b1
b2 =
=
= 1.2071
2
2
√
a3 = a2 b2 = 1.1892 · 1.2071 = 1.1981
a + b2
1.1892 · 1.2071
b3 = 2
=
= 1.1981
2
2
a4 = a3 b3 = 1.1981
a + b3
b4 = 3
= 1.1981
2
Thus,
√ AGM 1, 2 ≈ 1.198.
Let a1n = Hn1− ln n, where
Hn is the
1
1 nth harmonic number
+
+ ··· +
.
Let cn = +
n n+1 n+2
2n
1
1 1
(a) Calculate c1 , c2 , c3 , c4 .
Hn = 1 + + + · · · +
2 y3= x −1 over
n the interval [n, 2n] to prove that
(b) Use a comparison of rectangles with the area under
n+1
2n
dx 2n
dx
dx
1
(a) Show that an ≥ 0 for n ≥ 1. Hint: Show that
.
Hn ≥ 1
+ 1 ≤ cn x≤
+
x
2n
x
n
n
n
(b) Show that {an } is decreasing by interpreting an − an+1 as an area.
(c) Use
thelim
Squeeze
Theorem to determine lim cn .
an exists.
(c) Prove
that
n→∞
89.
n→∞
This limit, denoted γ , is known as Euler’s Constant. It appears in many areas of mathematics, including analysis and
number theory, and has been calculated to more than 100 million decimal places, but it is still not known whether γ is an
irrational number. The first 10 digits are γ ≈ 0.5772156649.
solution
(a) Since the function y = x1 is decreasing, the left endpoint approximation to the integral 1n+1 dx
x is greater than this
integral; that is,
1·1+
n+1
1
1
1
dx
· 1 + · 1 + ··· + · 1 ≥
2
3
n
x
1
or
Hn ≥
n+1
dx
.
x
1
y
1
1
1
2
3
1 2 3
1/n
n n+1
Moreover, since the function y = x1 is positive for x > 0, we have:
n+1
n
dx
dx
≥
.
x
1
1 x
May 23, 2011
x
S E C T I O N 10.2
Summing an Infinite Series
645
Thus,
Hn ≥
n
n
dx
= ln x = ln n − ln 1 = ln n,
1
1 x
and
an = Hn − ln n ≥ 0
for all n ≥ 1.
(b) To show that {an } is decreasing, we consider the difference an − an+1 :
an − an+1 = Hn − ln n − Hn+1 − ln(n + 1) = Hn − Hn+1 + ln(n + 1) − ln n
1
1
1
1
1
+ ln(n + 1) − ln n
= 1 + + ··· + − 1 + + ··· + +
2
n
2
n n+1
=−
1
+ ln(n + 1) − ln n.
n+1
n+1 dx
1
Now, ln(n + 1) − ln n = nn+1 dx
x , whereas n+1 is the right endpoint approximation to the integral n
x . Recalling
1
y = x is decreasing, it follows that
n+1
1
dx
≥
x
n
+
1
n
y
y=
1
x
1
n+1
n+1
n
x
so
an − an+1 ≥ 0.
(c) By parts (a) and (b), {an } is decreasing and 0 is a lower bound for this sequence. Hence 0 ≤ an ≤ a1 for all n. A
monotonic and bounded sequence is convergent, so limn→∞ an exists.
10.2 Summing an Infinite Series
Preliminary Questions
1. What role do partial sums play in defining the sum of an infinite series?
solution The sum of an infinite series is defined as the limit of the sequence of partial sums. If the limit of this sequence
does not exist, the series is said to diverge.
2. What is the sum of the following infinite series?
1 1
1
1
1
+ +
+
+
+ ···
4 8 16 32 64
solution This is a geometric series with c = 14 and r = 12 . The sum of the series is therefore
1
4
1
1
= 41 = .
1
2
1− 2
2
3. What happens if you apply the formula for the sum of a geometric series to the following series? Is the formula valid?
1 + 3 + 32 + 33 + 34 + · · ·
May 23, 2011
646
C H A P T E R 10
INFINITE SERIES
solution This is a geometric series with c = 1 and r = 3. Applying the formula for the sum of a geometric series
then gives
∞
3n =
n=0
1
1
=− .
1−3
2
Clearly, this is not valid: a series with all positive terms cannot have a negative sum. The formula is not valid in this case
because a geometric series with r = 3 diverges.
∞
1
1
4. Arvind asserts that
= 0 because 2 tends to zero. Is this valid reasoning?
n2
n
n=1
solution Arvind’s reasoning is not valid. Though the terms in the series do tend to zero, the general term in the
sequence of partial sums,
1
1
1
Sn = 1 + 2 + 2 + · · · + 2 ,
2
3
n
is clearly larger than 1. The sum of the series therefore cannot be zero.
∞
1
5. Colleen claims that
√ converges because
n
n=1
1
lim √ = 0
n→∞ n
Is this valid reasoning?
solution
Colleen’s reasoning is not valid. Although the general term of a convergent series must tend to zero, a series
∞
1
whose general term tends to zero need not converge. In the case of
√ , the series diverges even though its general
n
n=1
term tends to zero.
6. Find an N such that SN > 25 for the series
∞
2.
n=1
solution The Nth partial sum of the series is:
SN =
N
n=1
2 = 2 + · · · + 2 = 2N.
N
∞
7. Does there exist an N such that SN > 25 for the series
solution The series
∞
2−n ? Explain.
n=1
2−n is a convergent geometric series with the common ratio r =
n=1
S=
1
. The sum of the series is:
2
1
2
= 1.
1 − 12
Notice that the sequence of partial sums {SN } is increasing and converges to 1; therefore SN ≤ 1 for all N. Thus, there
does not exist an N such that SN > 25.
8. Give an example of a divergent infinite series whose general term tends to zero.
solution
Consider the series
∞
1
9
n=1 n 10
. The general term tends to zero, since lim
n→∞
1
9
n 10
= 0. However, the Nth partial
sum satisfies the following inequality:
SN =
1
9
1 10
+
1
1
9
2 10
1
+ ··· +
1
N
9
10
≥
N
N
9
10
9
1
= N 1− 10 = N 10 .
That is, SN ≥ N 10 for all N. Since lim N 10 = ∞, the sequence of partial sums Sn diverges; hence, the series
N→∞
diverges.
May 23, 2011
∞
1
9
n=1 n 10
S E C T I O N 10.2
Summing an Infinite Series
Exercises
1. Find a formula for the general term an (not the partial sum) of the infinite series.
1 1
1
1
1 5 25 125
(a) + +
+
+ ···
(b) + +
+
+ ···
3 9 27 81
1 2
4
8
1
22
33
44
−
+
−
+ ···
1 2·1 3·2·1 4·3·2·1
2
1
2
1
(d) 2
+ 2
+ 2
+ 2
+ ···
1 +1 2 +1 3 +1 4 +1
(c)
solution
(a) The denominators of the terms are powers of 3, starting with the first power. Hence, the general term is:
1
an = n .
3
(b) The numerators are powers of 5, and the denominators are the same powers of 2. The first term is a1 = 1 so,
an =
n−1
5
.
2
(c) The general term of this series is,
an = (−1)n+1
nn
.
n!
(d) Notice that the numerators of an equal 2 for odd values of n and 1 for even values of n. Thus,
⎧
2
⎪
⎪
⎨ n2 + 1 odd n
an =
⎪
⎪
⎩ 1
even n
n2 + 1
The formula can also be rewritten as follows:
n+1
1 + (−1) 2 +1
.
an =
n2 + 1
In Exercises
compute the
partial sums S2 , S4 , and S6 .
Write3–6,
in summation
notation:
1
1 1
1 +1 + · · ·
(a) 11+ +
3. 1 + 2 +4 2 9+ 216+ · · ·
2
3
4
1 1 1
(c) 1 − + − + · · ·
solution
3 5 7
1
5
125 625 3125 15,625
S2 =+1·+
+
+
+
·· 2 = 4;
(d)
2
9
16
25
36
(b)
1
1
1
1
+
+
+
+ ···
9 16 25 36
1
1
1
205
;
S4 = 1 + 2 + 2 + 2 =
144
2
3
4
1
1
1
1
1
5369
.
S6 = 1 + 2 + 2 + 2 + 2 + 2 =
3600
2
3
4
5
6
1
1
1
∞+
+
+ ···
1 · 2 (−1)
2 · 3k k −13 · 4
solution
k=1
5.
S2 =
1
1 1
4
2
1
+
= + = = ;
1·2 2·3
2 6
6
3
S4 = S2 + a3 + a4 =
2
1
1
2
1
1
4
+
+
= +
+
= ;
3 3·4 4·5
3 12 20
5
S6 = S4 + a5 + a6 =
1
1
4
1
1
6
4
+
+
= +
+
= .
5 5·6 6·7
5 30 42
7
∞
1
j!
j =1
May 23, 2011
647
648
C H A P T E R 10
INFINITE SERIES
2 3
7. The series S = 1 + 15 + 15 + 15 + · · · converges to 54 . Calculate SN for N = 1, 2, . . . until you find an SN
that approximates 54 with an error less than 0.0001.
solution
S1 = 1
S2 = 1 +
S3 = 1 +
S3 = 1 +
S4 = 1 +
S5 = 1 +
1
5
1
5
1
5
1
5
1
5
=
+
+
+
+
6
= 1.2
5
1
31
=
= 1.24
25
25
1
1
156
+
=
= 1.248
25 125
125
1
1
1
781
+
+
=
= 1.2496
25 125 625
625
1
1
1
1
3906
+
+
+
=
= 1.24992
25 125 625 3125
3125
Note that
1.25 − S5 = 1.25 − 1.24992 = 0.00008 < 0.0001
In Exercises 9 and 10, use
system to compute S10 , S100 , S500 , and S1000 for the series. Do these
1
1 algebra
1
1 a computer
−to the
+given
− value?
+ · · · is known to converge to e−1 (recall that 0! = 1). Calculate SN for
The series
S=
values suggest
convergence
0! 1! 2! 3!
N
=
1,
2,
.
.
.
until
you
find an SN that approximates e−1 with an error less than 0.001.
9.
1
1
1
1
π −3
=
−
+
−
+ ···
4
2 · 3 · 4 4 · 5 · 6 6 · 7 · 8 8 · 9 · 10
solution Write
an =
(−1)n+1
2n · (2n + 1) · (2n + 2)
Then
SN =
N
an
i=1
Computing, we find
π −3
≈ 0.0353981635
4
S10 ≈ 0.03535167962
S100 ≈ 0.03539810274
S500 ≈ 0.03539816290
S1000 ≈ 0.03539816334
It appears that SN → π−3
4 .
11. Calculate S3 , S4 , and S5 and then find the sum of the telescoping series
∞ 1
1
−
Sπ 4=
1
1
1
1 n++ 2 + · · ·
= 1 + n ++
90 n=1 24
34
44
solution
1 1
1 1
1 1
3
1 1
−
+
−
+
−
= − =
;
2 3
3 4
4 5
2 5
10
1 1
1 1
1
−
= − = ;
S4 = S3 +
5 6
2 6
3
1 1
5
1 1
−
= − =
.
S5 = S4 +
6 7
2 7
14
S3 =
May 23, 2011
S E C T I O N 10.2
Summing an Infinite Series
The general term in the sequence of partial sums is
1 1
1 1
1
1
1
1
1 1
SN =
−
+
−
+
−
+ ··· +
−
= −
;
2 3
3 4
4 5
N +1 N +2
2 N +2
thus,
1
1
−
N +2
N→∞ 2
S = lim SN = lim
N→∞
=
1
.
2
The sum of the telescoping series is therefore 12 .
∞
1
13. Calculate S3 , S4 , and S5 and then find the sum S =
using the identity
2−1
∞
4n
1
n=1
as a telescoping series
and
find
its
sum.
Write
1
1
1
1
n(n − 1)
−
=
n=3
2 2n − 1 2n + 1
4n2 − 1
solution
1 1
1 1 1
1 1 1
1
1
3
−
+
−
+
−
=
1−
= ;
1 3
2 3 5
2 5 7
2
7
7
1
1
4
1 1 1
−
=
1−
= ;
S4 = S3 +
2 7 9
2
9
9
1
1
1
5
1 1
−
=
1−
=
.
S5 = S4 +
2 9 11
2
11
11
1
S3 =
2
The general term in the sequence of partial sums is
1 1 1
1 1 1
1
1
1
1
1
1 1 1
SN =
−
+
−
+
−
+ ··· +
−
=
1−
;
2 1 3
2 3 5
2 5 7
2 2N − 1 2N + 1
2
2N + 1
thus,
1
1
1
1−
= .
2N + 1
2
N→∞ 2
S = lim SN = lim
N→∞
1
1
1
∞+ ··· .
+
+ 15. Find the sum of
1
1
·
3
3
·
5
5
Use partial fractions to rewrite · 7
as a telescoping series and find its sum.
n(n
+ 3)
solution We may write this sum asn=1
∞
n=1
∞
1
1
1
1
=
−
.
(2n − 1)(2n + 1)
2 2n − 1 2n + 1
n=1
The general term in the sequence of partial sums is
1 1 1
1 1 1
1
1
1
1
1
1 1 1
SN =
−
+
−
+
−
+ ··· +
−
=
1−
;
2 1 3
2 3 5
2 5 7
2 2N − 1 2N + 1
2
2N + 1
thus,
1
1
1
1−
= ,
2N + 1
2
N→∞ 2
lim SN = lim
N→∞
and
∞
n=1
1
1
= .
(2n − 1)(2n + 1)
2
In Exercises 17–22, use Theorem 3 to prove that the
∞ following series diverge.
(−1)n−1 and show that the series diverges.
Find a formula for the partial sum SN of
∞
n
n=1
17.
10n + 12
n=1
solution The general term,
n
, has limit
10n + 12
lim
n
n→∞ 10n + 12
= lim
Since the general term does not tend to zero, the series diverges.
May 23, 2011
1
n→∞ 10 + (12/n)
=
1
10
649
650
C H A P T E R 10
INFINITE SERIES
0 1 2 3
∞ + − + ···
−
1 2 3n 4
n−1
solution
general
term an = (−1)n−1 n−1
n2 +
1
n does not tend to zero. In fact, because limn→∞ n = 1, limn→∞ an
n=1 The
does not exist. By Theorem 3, we conclude that the given series diverges.
1
1
1
∞+ cos + cos + · · ·
21. cos 2 (−1)n3n2
4
1
solution
n=1 The general term an = cos n+1 tends to 1, not zero. By Theorem 3, we conclude that the given series
diverges.
19.
In Exercises
23–36, use the formula for the sum of a geometric series to find the sum or state that the series diverges.
∞ 2
1 +1−n
1 1 4n
23. +n=0 + 2 + · · ·
1 8 8
solution This is a geometric series with c = 1 and r = 18 , so its sum is
1
8
1
=
7/8
7
=
1 − 18
∞ −n
3 4
4
45
43
+
+
+ ···
11
n=353
55
54
solution Rewrite this series as
25.
∞ n
11
3
n=3
11
> 1, so it is divergent.
This is a geometric series with r =
3
∞
4 n
∞ −
27.
7 ·9(−3)n
n=−4
5n
n=2
4
solution This is a geometric series with c = 1 and r = − , starting at n = −4. Its sum is thus
9
c
1
95
59,049
cr −4
= 4
= 4
=
=
5
5
4 + 45
4 + 4
1−r
3328
r −r
9
·
4
5
4
9
9
∞
∞−n 29.
e
π n
n=1
e
n=0
solution Rewrite the series as
∞ n
1
e
n=1
to recognize it as a geometric series with c = 1e and r = 1e . Thus,
∞
e−n =
n=1
31.
1
e
1 − 1e
=
1
.
e−1
∞
8 + 2n
∞
5en3−2n
n=0
n=2
solution
Rewrite the series as
n ∞
∞ n
∞
∞ n
1
8
2
2
+
=
8
·
+
,
5n
5n
5
5
n=0
n=0
n=0
n=0
0
which is a sum of two geometric series. The first series has c = 8 15 = 8 and r = 15 ; the second has c = 25 = 1
and r = 25 . Thus,
n
∞
1
8
8
8·
=
= 4 = 10,
1
5
1− 5
5
n=0
0
∞ n
2
n=0
May 23, 2011
5
=
1
1 − 25
1
5
= 3 = ,
3
5
S E C T I O N 10.2
Summing an Infinite Series
651
and
∞
5
8 + 2n
35
= 10 + =
.
5n
3
3
n=0
5
5
5
∞+
33. 5 −
−n −3 5+n · · ·
4 3(−2)
42
4
8n
n=0 This is a geometric series with c = 5 and r = − 1 . Thus,
solution
4
5
1 n
5
5
=
5· −
=
= 5 = 4.
1
1
4
1
+
1 − −4
4
4
n=0
∞
7 49 343 2401
−3 +4
2−5
26 + · · ·
8 2 64+ 2 512
+ 3 4096
+ 4 + ···
2
7
7
7
7
solution This is a geometric series with c = 78 and r = − 78 . Thus,
35.
∞
7
7
7 n
7
7
8 = 8 =
· −
=
.
15
7
8
8
15
1 − −8
8
n=0
37. Which of the following are not geometric series?
∞ 25 n+ 5 + 1 + 3 + 9 + 27 + · · ·
97 3
5 25 125
(a)
29n
(c)
n=0
∞
n=0
n2
2n
(b)
(d)
∞
1
n4
n=3
∞
π −n
n=5
solution
(a)
∞ n
∞ n
7
7
7
=
: this is a geometric series with common ratio r =
.
29n
29
29
n=0
n=0
(b) The ratio between two successive terms is
1
n4
an+1
(n+1)4
=
=
=
1
an
(n + 1)4
4
n
This ratio is not constant since it depends on n. Hence, the series
4
n
.
n+1
∞
1
is not a geometric series.
n4
n=3
(c) The ratio between two successive terms is
(n+1)2
an+1
n+1
= 2 2
n
an
n
=
2
2n
1 2 1
(n + 1)2
·
=
1
+
· .
n
2
n2
2n+1
This ratio is not constant since it depends on n. Hence, the series
(d)
∞
n=5
π −n =
∞ n
1
n=5
π
∞ 2
n
n=0
2n
is not a geometric series.
: this is a geometric series with common ratio r =
1
.
π
∞
∞
∞
∞
an converges and
bn diverges,
then
(an + bn ) diverges. Hint: If not, derive a contradiction
39. Prove that if
1
diverges.
Use the method of Example 8 to show that
1/3
n=1
n=1
n=1
k
k=1
by writing
∞
n=1
May 23, 2011
bn =
∞
n=1
(an + bn ) −
∞
n=1
an
652
C H A P T E R 10
INFINITE SERIES
∞
∞
solution Suppose to the contrary that ∞
n=1 an converges,
n=1 bn diverges, but
n=1 (an + bn ) converges. Then
by the Linearity of Infinite Series, we have
∞
bn =
n=1
so that
∞
(an + bn ) −
n=1
∞
an
n=1
∞
n=1 bn converges, a contradiction.
∞ tonshown that each of the following statements is false.
Give a counterexample
9 +2
.
Prove the divergence of
∞
5n
to zero, then
an = 0.
(a) If the general term an tends n=0
41.
n=1
(b) The Nth partial sum of the infinite series defined by {an } is aN .
∞
(c) If an tends to zero, then
an converges.
n=1
∞
(d) If an tends to L, then
an = L.
n=1
solution
(a) Let an = 2−n . Then limn→∞ an = 0, but an is a geometric series with c = 20 = 1 and r = 1/2, so its sum is
1
= 2.
1 − (1/2)
(b) Let an = 1. Then the nth partial sum is a1 + a2 + · · · + an = n while an = 1.
∞
1
(c) Let an = √ . An example in the text shows that while an tends to zero, the sum
an does not converge.
n=1
n
∞
(d) Let an = 1. Then clearly an tends to L = 1, while the series n=1 an obviously diverges.
43. Compute the total area
of the (infinitely many) triangles in Figure 4.
∞
2
Suppose that S =
an is an infinite series with partial sum SN = 5 − 2 .
N
y
n=1
10
161
(a) What are the values of
an and 2 an ?
n=5
n=1
(b) What is the value of a3 ?
(c) Find a general formula for an .
∞
(d) Find the sum
an .
1 1
16 8
1
4
x
1
2
1
FIGURE 4
n=1
solution The area of a triangle with base B and height H is A = 12 BH . Because all of the triangles in Figure 4 have
height 12 , the area of each triangle equals one-quarter of the base. Now, for n ≥ 0, the nth triangle has a base which
1 to x = 1 . Thus,
extends from x = n+1
2n
2
1
1
1
B = n − n+1 = n+1
2
2
2
and
A=
1
1
B = n+3 .
4
2
The total area of the triangles is then given by the geometric series
∞
n=0
1
=
2n+3
∞
1
1
1 1 n
= 8 1 = .
8 2
4
1− 2
n=0
45. Find the total length of the infinite zigzag path in Figure 5 (each zag occurs at an angle of π4 ).
The winner of a lottery receives m dollars at the end of each year for N years. The present value (PV) of this prize
N
in today’s dollars is PV =
m(1 + r)−i , where r is the interest rate. Calculate PV if m = $50,000, r = 0.06, and
i=1
N = 20. What is PV if N = ∞?
π /4
π /4
1
FIGURE 5
May 23, 2011
S E C T I O N 10.2
Summing an Infinite Series
653
solution Because the angle at the lower left in Figure 5 has measure π4 and each zag in the path occurs at an angle of
π , every triangle in the figure is an isosceles right triangle. Accordingly, the length of each new segment in the path is
4
√1 times the length of the previous segment. Since the first segment has length 1, the total length of the path is
2
√
∞ √
1
2
1 n
=
= √
= 2 + 2.
√
1
√
2
2−1
1−
n=0
2
∞
1 integer, then
47. Show that if a
is a positive
Evaluate
. Hint: Find constants A, B, and C such that
n(n + 1)(n + 2) ∞
n=1
1
1
1C
11
= = 1A++ +B· · · +
+ a)+ 2)a n 2 n + 1 + an + 2
n(nn(n
+ 1)(n
n=1
solution
By partial fraction decomposition
1
A
B
= +
;
n (n + a)
n
n+a
clearing the denominators gives
1 = A(n + a) + Bn.
Setting n = 0 then yields A = a1 , while setting n = −a yields B = − a1 . Thus,
1
1
1
1
= a − a =
n (n + a)
n
n+a
a
1
1
−
n n+a
,
and
∞
n=1
∞
1
1 1
1
=
−
.
n(n + a)
a n n+a
n=1
For N > a, the Nth partial sum is
1
1 1
1
1
1
1
1
1
1 + + + ··· +
−
+
+
+ ··· +
.
SN =
a
2 3
a
a N +1 N +2 N +3
N +a
Thus,
1
1
1 1
1
= lim SN =
1 + + + ··· +
.
n(n + a) N→∞
a
2 3
a
n=1
∞
49. LetA{bball
a
sequence
and
let
a
=
b
−
b
.
Show
that
andground,
only ifit lim
bn to
exists.
n } be
n
n
n converges
n−1
dropped from a height of 10 ft begins to bounce. Eachatime
it strikesifthe
returns
two-thirds of
n→∞
its previous height. What is the total distance traveled by then=1
ball if it bounces infinitely many
times?
∞
solution Let an = bn − bn−1 . The general term in the sequence of partial sums for the series
an is then
∞
n=1
SN = (b1 − b0 ) + (b2 − b1 ) + (b3 − b2 ) + · · · + (bN − bN−1 ) = bN − b0 .
Now, if lim bN exists, then so does lim SN and
N→∞
N→∞
∞
an converges. On the other hand, if
n=1
lim SN exists, which implies that lim bN also exists. Thus,
N→∞
N→∞
∞
∞
an converges, then
n=1
an converges if and only if lim bn exists.
n→∞
n=1
Assumptions
Show, by giving counterexamples, that the assertions of Theorem 1 are not valid if the
Further
Insights Matter
and Challenges
∞
∞
n=0
n=0
use the
Exercises 51–53
formula
an and
bn are not convergent.
series
1 + r + r 2 + · · · + r N−1 =
1 − rN
1−r
7
51. Professor GeorgeAndrews of Pennsylvania State University observed that we can use Eq. (7) to calculate the derivative
of f (x) = x N (for N ≥ 0). Assume that a = 0 and let x = ra. Show that
f (a) = lim
x→a
and evaluate the limit.
May 23, 2011
x N − aN
rN − 1
= a N−1 lim
x−a
r→1 r − 1
654
C H A P T E R 10
INFINITE SERIES
solution According to the definition of derivative of f (x) at x = a
x N − aN
.
x→a x − a
f (a) = lim
Now, let x = ra. Then x → a if and only if r → 1, and
N − aN
N − aN
aN r N − 1
x
rN − 1
(ra)
= lim
= lim
= a N−1 lim
.
f (a) = lim
x→a x − a
ra − a
r→1
r→1 a (r − 1)
r→1 r − 1
By Eq. (7) for a geometric sum,
rN − 1
1 − rN
=
= 1 + r + r 2 + · · · + r N−1 ,
1−r
r −1
so
rN − 1
= lim 1 + r + r 2 + · · · + r N−1 = 1 + 1 + 12 + · · · + 1N−1 = N.
r→1 r − 1
r→1
lim
Therefore, f (a) = a N−1 · N = N a N−1
53. Verify the Gregory–Leibniz formula as follows.
de2 in
Fermat
used
tothat
compute the area under the graph of f (x) = x N over [0, A]. For
(a) Set Pierre
r = −x
Eq. (7)
andgeometric
rearrange series
to show
0 < r < 1, let F (r) be the sum of the areas of the infinitely many right-endpoint rectangles with endpoints Ar n , as
in Figure 6. As r tends to 1,1 the rectangles
become narrower and F (r) tends (−1)
to theNarea
x 2N under the graph.
=1 −
1−
x 2 + x 4 − · · · + (−1)N−1 x 2N−2 +
r
2
1 N+1
+x
1 + x2
(a) Show that F (r) = A
.
1 − r N+1
1],
(b) Show, by integrating over [0,
A that
x N dx = lim F (r).
(b) Use Eq. (7) to evaluate
1 2N
1 1 r→1
x dx
1
(−1)N−1
π
0
= 1 − + − + ··· +
+ (−1)N
4
3 5 7
2N − 1
0 1 + x2
(c) Use the Comparison Theorem for integrals to prove that
1 2N
1
x dx
≤
0≤
2N + 1
0 1 + x2
Hint: Observe that the integrand is ≤ x 2N .
(d) Prove that
1 1 1 1
π
= 1 − + − + − ···
4
3 5 7 9
1 , and thereby conclude that
Hint: Use (b) and (c) to show that the partial sums SN of satisfy SN − π4 ≤ 2N+1
π
lim SN = 4 .
N→∞
solution
(a) Start with Eq. (7), and substitute −x 2 for r:
1 + r + r 2 + · · · + r N−1 =
1 − rN
1−r
1 − x 2 + x 4 + · · · + (−1)N−1 x 2N−2 =
1 − x 2 + x 4 + · · · + (−1)N−1 x 2N−2 =
1 − (−1)N x 2N
1 − (−x 2 )
1
(−1)N x 2N
−
1 + x2
1 + x2
(−1)N x 2N
1
= 1 − x 2 + x 4 + · · · + (−1)N−1 x 2N−2 +
2
1+x
1 + x2
(b) The integrals of both sides must be equal. Now,
1
1
−1 x = tan−1 1 − tan−1 0 = π
dx
=
tan
2
4
0 1+x
0
1
while
1
0
(−1)N x 2N
1 − x 2 + x 4 + · · · + (−1)N−1 x 2N−2 +
1 + x2
May 23, 2011
dx
S E C T I O N 10.2
Summing an Infinite Series
655
1 2N
1
1
1
x dx
= x − x 3 + x 5 + · · · + (−1)N−1
x 2N−1 + (−1)N
3
5
2N − 1
0 1 + x2
1 2N
1
x dx
1 1
+ (−1)N
= 1 − + + · · · + (−1)N−1
3 5
2N − 1
0 1 + x2
(c) Note that for x ∈ [0, 1], we have 1 + x 2 ≥ 1, so that
0≤
x 2N
≤ x 2N
1 + x2
By the Comparison Theorem for integrals, we then see that
1
1
1 2N
1
1
x dx
2N
2N+1
=
x
≤
x dx =
0≤
2
2N
+
1
2N
+1
1
+
x
0
0
0
(d) Write
an = (−1)n
1
,
2n − 1
n≥1
and let SN be the partial sums. Then
1 2N
1 x 2N dx
π 1
x dx N
≤
=
SN − = (−1)
2
2
4
2N
+1
1
+
x
1
+
x
0
0
Thus limN→∞ SN =
π
so that
4
π
1 1 1 1
= 1 − + − + − ...
4
3 5 7 9
55. The Koch snowflake (described in 1904 by Swedish mathematician Helge von Koch) is an infinitely jagged “fractal”
Cantor’sasDisappearing
Table (following
Knop of but
Hamilton
Take
table
of length
(Figure
curve obtained
a limit of polygonal
curves (it Larry
is continuous
has noCollege)
tangent line
at aany
point).
BeginL with
an
7). At stage
1, remove
section
of length
centered at
theedge
midpoint.
Twoedges
sections
remain, each
with length
less
equilateral
triangle
(stage 0)the
and
produce
stage 1L/4
by replacing
each
with
four
of
one-third
the
length,
arranged
2
L/2.
stage 2,the
remove
sections
length
L/4
fromeach
eachedge
of these
stage removes
L/8 of the
as inthan
Figure
8. At
Continue
process:
At theofnth
stage,
replace
withtwo
foursections
edges of(this
one-third
the length.
table). Now four sections remain, each of length less than L/4. At stage 3, remove
the four central sections of length
4
(a) Show
that the perimeter Pn of the polygon at the nth stage satisfies Pn = 3 Pn−1 . Prove that lim Pn = ∞. The
n→∞
L/43 , etc.
snowflake has infinite length.
(a) Show that at the N th stage, each remaining section has length lessn−1
than L/2N and that the total amount of table
new triangles are added at the nth stage,
(b) Let A0 be the area of the original equilateral triangle. Show that (3)4
removed is
8
each with area A0 /9n (for n ≥ 1). Show that the
is 5 A0 .
total area of the Koch snowflake
1
1
1 1
+ +
+ · · · + N+1
L
4 8 16
2
(b) Show that in the limit as N → ∞, precisely one-half of the table remains.
This result is curious, because there are no nonzero intervals of table left (at each stage, the remaining sections have
Stage 1
Stage 2
Stageplace
3
a length less than L/2N ). So the table has “disappeared.”
However,
we can
any object longer than L/4 on the
8 of the removed sections.
table. It will not fall through because it will not fitFIGURE
through any
solution
(a) Each edge of the polygon at the (n − 1)st stage is replaced by four edges of one-third the length; hence the perimeter
of the polygon at the nth stage is 43 times the perimeter of the polygon at the (n − 1)th stage. That is, Pn = 43 Pn−1 . Thus,
P1 =
4
P0 ;
3
P2 =
4
P1 =
3
2
4
P0 ,
3
n
and, in general, Pn = 43 P0 . As n → ∞, it follows that
P3 =
4
P2 =
3
3
4
P0 ,
3
n
4
= ∞.
n→∞ 3
lim Pn = P0 lim
n→∞
(b) When each edge is replaced by four edges of one-third the length, one new triangle is created. At the (n − 1)st stage,
there are 3 · 4n−1 edges in the snowflake, so 3 · 4n−1 new triangles are generated at the nth stage. Because the area of an
equilateral triangle is proportional to the square of its side length and the side length for each new triangle is one-third
the side length of triangles from the previous stage, it follows that the area of the triangles added at each stage is reduced
by a factor of 19 from the area of the triangles added at the previous stage. Thus, each triangle added at the nth stage has
an area of A0 /9n . This means that the nth stage contributes
n
A
3
4
3 · 4n−1 · n0 = A0
9
4
9
May 23, 2011
656
C H A P T E R 10
INFINITE SERIES
to the area of the snowflake. The total area is therefore
∞ n
4
8
3
3
3
4
4
A = A0 + A0
= A0 + A0 9 4 = A0 + A0 · = A0 .
4
9
4 1−
4
5
5
9
n=1
10.3 Convergence of Series with Positive Terms
Preliminary Questions
1. Let S =
∞
an . If the partial sums SN are increasing, then (choose the correct conclusion):
n=1
(a) {an } is an increasing sequence.
(b) {an } is a positive sequence.
solution The correct response is (b). Recall that SN = a1 + a2 + a3 + · · · + aN ; thus, SN − SN−1 = aN . If SN is
increasing, then SN − SN−1 ≥ 0. It then follows that aN ≥ 0; that is, {an } is a positive sequence.
2. What are the hypotheses of the Integral Test?
solution The hypotheses for the Integral Test are: A function f (x) such that an = f (n) must be positive, decreasing,
and continuous for x ≥ 1.
∞
n−3.2 converges?
3. Which test would you use to determine whether
solution
converges.
Because n−3.2 =
n=1
1 , we see that the indicated series is a p-series with p = 3.2 > 1. Therefore, the series
n3.2
∞
1
4. Which test would you use to determine whether
√
n
2 + n
n=1
converges?
solution
Because
1
1
√ < n =
2
2n + n
n
1
,
2
and
∞ n
1
n=1
2
is a convergent geometric series, the comparison test would be an appropriate choice to establish that the given series
converges.
5. Ralph hopes to investigate the convergence of
∞ −n
∞
e
1
by comparing it with
. Is Ralph on the right track?
n
n
n=1
solution
n=1
No, Ralph is not on the right track. For n ≥ 1,
e−n
1
< ;
n
n
∞
1
is a divergent series. The Comparison Test therefore does not allow us to draw a conclusion about the
n
n=1
∞ −n
e
convergence or divergence of the series
.
n
however,
n=1
Exercises
In Exercises 1–14, use the Integral Test to determine whether the infinite series is convergent.
1.
∞
1
n4
n=1
1
Let f (x) = 4 . This function is continuous, positive and decreasing on the interval x ≥ 1, so the Integral
x
Test applies. Moreover,
solution
May 23, 2011
S E C T I O N 10.3
Convergence of Series with Positive Terms
657
∞
R
1
1
dx
−4 dx = − 1 lim
=
lim
x
−
1
= .
3 R→∞ R 3
3
R→∞ 1
x4
1
The integral converges; hence, the series
∞
1
also converges.
n4
n=1
3.
∞
∞−1/3
n
1
n=1
n+3
n=1
1
1
Let f (x) = x − 3 = √
. This function is continuous, positive and decreasing on the interval x ≥ 1, so the
3
x
Integral Test applies. Moreover,
R
∞
3
lim R 2/3 − 1 = ∞.
x −1/3 dx = lim
x −1/3 dx =
2 R→∞
R→∞ 1
1
solution
The integral diverges; hence, the series
∞
n−1/3 also diverges.
n=1
∞
n2
∞
5.
1
3 + 9)5/2
(n√
n=25
n
−4
n=5
solution
Let f (x) = x2
5/2 . This function is positive and continuous for x ≥ 25. Moreover, because
x3 + 9
f (x) =
2x(x 3 + 9)
5/2
3/2
− x 2 · 52 (x 3 + 9)
· 3x 2
(x 3 + 9)
5
=
x(36 − 11x 3 )
2(x 3 + 9)
7/2
,
we see that f (x) < 0 for x ≥ 25, so f is decreasing on the interval x ≥ 25. The Integral Test therefore applies. To
evaluate the improper integral, we use the substitution u = x 3 + 9, du = 3x 2 dx. We then find
∞
R
R 3 +9
1
x2
x2
du
lim
dx
=
lim
dx
=
3 R→∞ 15634 u5/2
R→∞ 25 (x 3 + 9)5/2
25 (x 3 + 9)5/2
2
1
1
2
= − lim
−
.
=
9 R→∞ (R 3 + 9)3/2
156343/2
9 · 156343/2
The integral converges; hence, the series
7.
∞
n2
5/2
3
n=25 n + 9
also converges.
∞
∞ 1
n
n2 + 12
n=1
(n + 1)3/5
n=1
1
Let f (x) = 2
. This function is positive, decreasing and continuous on the interval x ≥ 1, hence the
x +1
Integral Test applies. Moreover,
R
∞
dx
dx
−1 R − π = π − π = π .
tan
=
lim
=
lim
4
2
4
4
R→∞ 1 x 2 + 1
R→∞
1 x2 + 1
solution
The integral converges; hence, the series
9.
∞
1
n=1
n2 + 1
also converges.
∞
∞ 1
1
n(n + 1)
n=1
n2 − 1
n=4
1
. This function is positive, continuous and decreasing on the interval x ≥ 1, so the
x(x + 1)
Integral Test applies. We compute the improper integral using partial fractions:
R
∞
1
1
1
x R
1
R
dx
= lim
−
dx = lim ln
−
ln
= ln 1 − ln = ln 2.
=
lim
ln
x
x+1
x + 1 1
R+1
2
2
R→∞ 1
R→∞
R→∞
1 x(x + 1)
solution
Let f (x) =
The integral converges; hence, the series
∞
n=1
May 23, 2011
1
converges.
n(n + 1)
658
C H A P T E R 10
INFINITE SERIES
11.
∞
∞ 1
2
−n
2
n(lnnen)
n=2
n=1
solution
Let f (x) =
1
x(ln x)2
f (x) = −
. This function is positive and continuous for x ≥ 2. Moreover,
1
x 2 (ln x)4
1 · (ln x)2 + x · 2 (ln x) ·
1
x
=−
1
(ln x)2 + 2 ln x .
x 2 (ln x)4
Since ln x > 0 for x > 1, f (x) is negative for x > 1; hence, f is decreasing for x ≥ 2. To compute the improper integral,
1
we make the substitution u = ln x, du = dx. We obtain:
x
∞
R
ln R
1
1
du
dx
=
lim
dx
=
lim
2
2
R→∞
R→∞
x(ln
x)
x(ln
x)
2
2
ln 2 u2
1
1
1
−
=
.
= − lim
ln 2
ln 2
R→∞ ln R
The integral converges; hence, the series
∞
n=2
1
also converges.
n(ln n)2
∞
∞1
13.
ln n
2ln n 2
n=1
n
n=1
solution Note that
2ln n = (eln 2 )ln n = (eln n )ln 2 = nln 2 .
Thus,
∞
1
n=1
2ln n
=
∞
n=1
1
.
nln 2
1
. This function is positive, continuous and decreasing on the interval x ≥ 1; therefore, the Integral
x ln 2
Test applies. Moreover,
∞
R
dx
dx
1
lim (R 1−ln 2 − 1) = ∞,
= lim
=
1 − ln 2 R→∞
R→∞ 1 x ln 2
1 x ln 2
Now, let f (x) =
because 1 − ln 2 > 0. The integral diverges; hence, the series
∞
n=1
1
also diverges.
2ln n
∞
∞
1
n−3 .
converges
by
using
the
Comparison
Test
with
3 + 8n
n
n
n=1
3lnn=1
∞
n=1
solution We compare the series with the p-series
n−3 . For n ≥ 1,
∞
15. Show
that1
n=1
1
1
≤ 3.
n3 + 8n
n
Since
∞
∞
1
1
converges
(it
is
a
p-series
with
p
=
3
>
1),
the
series
also converges by the Comparison Test.
n3
n3 + 8n
n=1
n=1
∞
∞1
∞
17. Let S =
√ . 1Verify that for n ≥ 1,
n−1 .
diverges by comparing with
Show thatn + n
n=1
2−3
n
n=2
n=2 1
1
1
1
√ ≤ ,
√ ≤ √
n
n+ n
n+ n
n
1
1
Can either inequality be used to show that S diverges? Show that
and conclude that S diverges.
√ ≥
2n
n+ n
√
√
√
solution For n ≥ 1, n + n ≥ n and n + n ≥ n. Taking the reciprocal of each of these inequalities yields
1
1
√ ≤
n
n+ n
May 23, 2011
and
1
1
√ ≤ √ .
n
n+ n
S E C T I O N 10.3
These inequalities indicate that the series
∞
n=1
Convergence of Series with Positive Terms
659
∞
∞
∞
1
1
1
1
and
and
√ is smaller than both
√ ; however,
n
n
n+ n
n
∞
1
√ both diverge so neither inequality allows us to show that S diverges.
n
n=1
√
√
On the other hand, for n ≥ 1, n ≥ n, so 2n ≥ n + n and
n=1
n=1
n=1
1
1
.
√ ≥
2n
n+ n
The series
∞
∞
1
1
=2
diverges, since the harmonic series diverges. The Comparison Test then lets us conclude
2n
n
n=1
n=1
that the larger series
∞
n=1
1
√ also diverges.
n+ n
In Exercises 19–30, use the Comparison Test to determine whether the infinite series
is
∞convergent.
1
Which
of
the
following
inequalities
can
be
used
to
study
the
convergence
of
√ ? Explain.
∞
2
1
n + n
n=2
19.
n2n
n=1
1
1
1
1 ,n
∞√
√ ≤
√ ≤ 2
2 + n n1
2+ n
n
n
n
solution We compare with the geometric series
. For n ≥ 1,
2
n=1
1
1
≤ n =
n2n
2
Since
∞ n
1
2
n=1
n
1
.
2
converges (it is a geometric series with r = 12 ), we conclude by the Comparison Test that
∞
1
also
n2n
n=1
converges.
∞
∞ 1
n3
n1/3 5+ 2n
n=1
n + 4n + 1
n=1
solution For n ≥ 1,
21.
1
1
≤ n
2
n1/3 + 2n
The series
∞
1
1
1
,
so
it
converges.
By
the
Comparison
test,
so
does
is
a
geometric
series
with
r
=
.
n=1 2n
2
n1/3 + 2n
∞
n=1
∞
∞ 4
23.
1
m! + 4m
m=1
3
n=1 n + 2n − 1
solution For m ≥ 1,
4
4
≤ m =
m
m! + 4
4
The series
m−1
1
.
4
∞ m−1
1
1
is a geometric series with r = , so it converges. By the Comparison Test we can therefore
4
4
m=1
conclude that the series
∞
m=1
4
also converges.
m! + 4m
∞
sin2 √
k
∞
25.
n
k2
k=1
n−3
n=4
solution For k ≥ 1, 0 ≤ sin2 k ≤ 1, so
0≤
The series
1
sin2 k
≤ 2.
2
k
k
∞
1
is a p-series with p = 2 > 1, so it converges. By the Comparison Test we can therefore conclude that
k2
k=1
∞
sin2 k
also converges.
the series
k2
k=1
May 23, 2011
660
C H A P T E R 10
INFINITE SERIES
27.
∞
∞ 2 1/3
k
3n + 3−n
5/4
n=1
k
−k
k=2
Since 3−n > 0 for all n,
solution
n
2
2
1
≤
=
2
.
3n + 3−n
3n
3
The series
∞ n
1
1
2
is a geometric series with r = , so it converges. By the Comparison Theorem we can therefore
3
3
n=1
conclude that the series
∞
n=1
29.
2
also converges.
3n + 3−n
∞
∞ 1
2
(n +
1)!
2−k
n=1
k=1
Note that for n ≥ 2,
solution
(n + 1)! = 1 · 2 · 3 · · · n · (n + 1) ≤ 2n
n factors
so that
∞
n=1
But
∞
∞
n=2
n=2
1
1
1
=1+
≤1+
(n + 1)!
(n + 1)!
2n
∞
1
1
1
converges as
n=2 2n is a geometric series with ratio r = 2 , so it converges. By the comparison test,
(n + 1)!
∞
n=1
well.
Exercise
31–36: For all a > 0 and b > 1, the inequalities
∞
n!
n=1
ln n ≤ na ,
n3
na < bn
are true for n sufficiently large (this can be proved using L’Hopital’s Rule). Use this, together with the Comparison
Theorem, to determine whether the series converges or diverges.
31.
∞
ln n
n3
n=1
solution
For n sufficiently large (say n = k, although in this case n = 1 suffices), we have ln n ≤ n, so that
∞
∞
∞
ln n
n
1
≤
=
3
3
n
n
n2
n=k
n=k
This is a p-series with p = 2 > 1, so it converges. Thus
terms for 1 ≤ n ≤ k does not affect this result.
33.
n=k
∞
ln n
n=k n3 also converges; adding back in the finite number of
∞
(ln n)100
∞
1
n1.1
n=1
ln m
m=2
solution
Choose N so that ln n ≤ n0.0005 for n ≥ N . Then also for n > N, (ln n)100 ≤ (n0.0005 )100 = n0.05 . Then
∞
∞
∞
(ln n)100
n0.05
1
≤
=
1.1
1.1
1.05
n
n
n
n=N
But
∞
n=N
n=N
n=N
1
(ln n)1 00
is a p-series with p = 1.05 > 1, so is convergent. It follows that ∞
is also convergent;
n=N
n1.1
n1.05
adding back in the finite number of terms for n = 1, 2, . . . , N − 1 shows that
∞
(ln n)100
converges as well.
n1.1
n=1
∞
n=1
1
(ln n)10
May 23, 2011
Convergence of Series with Positive Terms
S E C T I O N 10.3
35.
661
∞
n
3n
n=1
Choose N such that n ≤ 2n for n ≥ N . Then
solution
∞
∞ n
n
2
≤
n
3
3
n=N
n=N
2
< 1, so it converges. Thus the series on the left converges as well. Adding
3
∞
n
converges.
back in the finite number of terms for n < N shows that
3n
The latter sum is a geometric series with r =
n=1
∞
1
sin 2 converges. Hint: Use sin x ≤ x for x ≥ 0.
n
∞ 5
37. Show
that
n
2n n=1
n=1
solution For n ≥ 1,
1
0 ≤ 2 ≤ 1 < π;
n
therefore, sin 12 > 0 for n ≥ 1. Moreover, for n ≥ 1,
n
1
1
sin 2 ≤ 2 .
n
n
∞
1
is a p-series with p = 2 > 1, so it converges. By the Comparison Test we can therefore conclude that
n2
The series
n=1
∞
the series
n=1
1
sin 2 also converges.
n
In Exercises 39–48,
use the Limit Comparison Test to prove convergence or divergence of the infinite series.
∞
sin(1/n)
Does
converge?
∞
2
n
ln n
n=2
39.
n4 − 1
n=2
n2
1
1
n2
n2
Let an = 4
. For large n, 4
≈ 4 = 2 , so we apply the Limit Comparison Test with bn = 2 .
n −1
n −1
n
n
n
solution
We find
n2
4
an
n4
L = lim
= lim n 1−1 = lim 4
= 1.
n→∞ bn
n→∞
n→∞ n − 1
2
n
The series
∞
n=1
∞
1
1
is a p-series with p = 2 > 1, so it converges; hence,
also converges. Because L exists, by the
n2
n2
Limit Comparison Test we can conclude that the series
∞
n=2
n=2
n2
converges.
4
n −1
∞
∞ n
41.
1
3
√
n=2 nn2+−1 n
n=2
solution
n
1
n
n
Let an = . For large n, ≈ √ = √ , so we apply the Limit Comparison test with
3
3
3
n
n
n +1
n +1
1
bn = √ . We find
n
an
= lim
L = lim
n→∞ bn
n→∞
The series
√
√n
n3
n3 +1
= lim = 1.
1
n→∞
√
n3 + 1
n
∞
∞
1
1
√ is a p-series with p = 12 < 1, so it diverges; hence,
√ also diverges. Because L > 0, by the
n
n
n=1
Limit Comparison Test we can conclude that the series
∞
n=2
May 23, 2011
n=2
n
diverges.
n3 + 1
662
C H A P T E R 10
INFINITE SERIES
43.
∞
∞ 3n + 53
n
n(n− 1)(n − 2)
n=3
7
2
n=2 n + 2n + 1
Let an =
solution
3n + 5
3
3n + 5
3n
. For large n,
≈ 3 = 2 , so we apply the Limit Comparison
n(n − 1)(n − 2)
n(n − 1)(n − 2)
n
n
1
Test with bn = 2 . We find
n
3n+5
an
3n3 + 5n2
n(n+1)(n+2)
= lim
= lim
= 3.
1
n→∞ bn
n→∞
n→∞ n(n + 1)(n + 2)
2
L = lim
n
The series
∞
n=1
∞
1
1
is
a
p-series
with
p
=
2
>
1,
so
it
converges;
hence,
the
series
also converges. Because L
n2
n2
exists, by the Limit Comparison Test we can conclude that the series
∞
n=3
n=3
3n + 5
converges.
n(n − 1)(n − 2)
∞
∞ 1n
45.
√ e +n
n+
ln n
n=1
e2n − n2
n=1
solution Let
1
an = √
n + ln n
For large n,
√
n + ln n ≈
√
1
n, so apply the Comparison Test with bn = √ . We find
n
√
an
1
n
1
= lim
= lim √
=1
·
L = lim
n→∞ bn
n→∞ n + ln n
n→∞
1
√n
1 + ln
n
The series
∞
1
1
√ is a p-series with p = < 1, so it diverges. Because L exists, the Limit Comparison Test tells us the
2
n
n=1
the original series also diverges.
∞ ∞
1
∞
47.
n−2 .
1−
cos
ln(n + 4) Hint: Compare with
n
5/2
n=1
n=1
n
n=1
1
1
solution Let an = 1 − cos , and apply the Limit Comparison Test with bn = 2 . We find
n
n
− 12 sin x1
1 − cos n1
1 − cos x1
sin x1
1
an
x
lim
= lim
=
lim
=
lim
=
.
1
1
n→∞ bn
n→∞
x→∞
x→∞
2 x→∞ 1
− 23
2
2
L = lim
n
x
x
x
As x → ∞, u = x1 → 0, so
L=
sin x1
1
1
1
sin u
lim
= .
= lim
2 x→∞ 1
2 u→0 u
2
x
∞
1
is a p-series with p = 2 > 1, so it converges. Because L exists, by the Limit Comparison Test we can
n2
n=1
∞ 1
also converges.
1 − cos
conclude that the series
n
The series
n=1
In Exercises
∞ 49–78, determine convergence or divergence using any method covered so far.
(1 − 2−1/n ) Hint: Compare with the harmonic series.
∞
1
n=1
49.
n2 − 9
n=4
1
1
solution Apply the Limit Comparison Test with an = 2
and bn = 2 :
n −9
n
1
2
an
n2
= lim n 1−9 = lim 2
= 1.
n→∞ bn
n→∞
n→∞ n − 9
2
L = lim
n
May 23, 2011
S E C T I O N 10.3
Since the p-series
Convergence of Series with Positive Terms
663
∞
∞
1
1
converges,
the
series
also converges. Because L exists, by the Limit Comparison Test
n2
n2
n=1
we can conclude that the series
√
∞
∞ n 2
51.
cos n
4n + 9
n=1
n2
∞
n=4
n=4
1
converges.
2
n −9
√
n=1
solution Apply the Limit Comparison Test with an =
1
n
and bn = √ :
4n + 9
n
√
n
1
an
n
= .
= lim 4n+9
= lim
L = lim
n→∞ bn
n→∞ √1
n→∞ 4n + 9
4
n
∞
1
The series
√ is a divergent p-series. Because L > 0, by the Limit Comparison Test we can conclude that the series
n
n=1
√
∞
n
also diverges.
4n + 9
n=1
53.
∞ 2
n −n
∞
n − cos n
n5 + n 3
n=1
n
n=1
solution
n (n − 1)
n2 − n
n−1
=
= 4
First rewrite an = 5
and observe
n +n
n n +1
n4 + 1
1
n
n−1
< 4 = 3
n4 + 1
n
n
for n ≥ 1. The series
∞
1
is a convergent p-series, so by the Comparison Test we can conclude that the series
n3
n=1
∞ 2
n −n
also converges.
n5 + n
n=1
55.
∞
∞
(4/5)−n 1
n=5
n2 + sin n
n=1
solution
∞ −n
4
n=5
5
=
∞ n
5
n=5
which is a geometric series starting at n = 5 with ratio r =
∞
∞ 1
1
n3/2 ln2 n
n=2
n
n=1 3
solution For n ≥ 3, ln n > 1, so n3/2 ln n > n3/2 and
4
5
> 1. Thus the series diverges.
4
57.
1
n3/2 ln n
The series
1
< 3/2 .
n
∞
∞
1
1
is a convergent p-series, so the series
also converges. By the Comparison Test we can
3/2
n
n3/2
n=1
therefore conclude that the series
∞
n=3
n=3
∞
1
1
converges.
Hence,
the
series
also converges.
3/2
3/2
n ln n
n ln n
n=2
∞
∞1/k
59.
4 (ln n)12
k=1
n=2
n9/8
solution
lim ak = lim 41/k = 40 = 1 = 0;
k→∞
therefore, the series
∞
k→∞
41/k diverges by the Divergence Test.
k=1
May 23, 2011
664
C H A P T E R 10
INFINITE SERIES
61.
∞
∞ 1
4n
(ln n)n4
n=2
5 − 2n
n=1
solution By the comment preceding Exercise 31, we can choose N so that for n ≥ N, we have ln n < n1/8 , so that
(ln n)4 < n1/2 . Then
∞
n=N
∞
1
1
>
(ln n)4
n1/2
n=N
which is a divergent p-series. Thus the series on the left diverges as well, and adding back in the finite number of terms
∞
1
diverges.
for n < N does not affect the result. Thus
(ln n)4
n=2
63.
∞
∞ 1 n
2
n ln nn − n
n=1
3 −n
n=1
solution
For n ≥ 2, n ln n − n ≤ n ln n; therefore,
1
1
≥
.
n ln n − n
n ln n
1
. For x ≥ 2, this function is continuous, positive and decreasing, so the Integral Test applies. Using
x ln x
the substitution u = ln x, du = x1 dx, we find
Now, let f (x) =
∞
R
ln R
dx
dx
du
= lim
= lim
= lim (ln(ln R) − ln(ln 2)) = ∞.
x
ln
x
x
ln
x
u
R→∞ 2
R→∞ ln 2
R→∞
2
The integral diverges; hence, the series
the series
∞
n=2
65.
∞
n=2
1
also diverges. By the Comparison Test we can therefore conclude that
n ln n
1
diverges.
n ln n − n
∞
∞1
nn
1
n=1
n(ln n)2 − n
n=1
For n ≥ 2, nn ≥ 2n ; therefore,
solution
1
1
≤ n =
nn
2
The series
∞ n
1
n=1
2
∞ n
1
is a convergent geometric series, so
therefore conclude that the series
n
1
.
2
n=2
2
∞
∞
1
1
converges.
Hence,
the
series
converges.
nn
nn
n=2
67.
also converges. By the Comparison Test we can
n=1
∞
n
1 + (−1)
∞
n2 − 4n3/2
n
n=1
n3
n=1
solution
Let
an =
1 + (−1)n
n
Then
an =
0
2
1
2k = k
n odd
n = 2k even
Therefore, {an } consists of 0s in the odd places
and the harmonic series in the even places, so
the harmonic series, which diverges. Thus ∞
i=1 an diverges as well.
∞
2 + (−1)n
n3/2
n=1
May 23, 2011
∞
i=1 an is just the sum of
Convergence of Series with Positive Terms
S E C T I O N 10.3
69.
∞
sin
n=1
665
1
n
solution Apply the Limit Comparison Test with an = sin
L = lim
sin n1
n→∞
1
n
1
1
and bn = :
n
n
= lim
u→0
sin u
= 1,
u
where u = n1 . The harmonic series diverges. Because L > 0, by the Limit Comparison Test we can conclude that the
∞
1
sin also diverges.
series
n
n=1
71.
∞
2n + 1
∞
sin(1/n)
4n √
n=1
n
n=1
solution
For n ≥ 3, 2n + 1 < 2n , so
2n + 1
2n
< n =
n
4
4
The series
∞ n
1
n=1
2
is a convergent geometric series, so
∞ n
1
n=3
therefore conclude that the series
n
1
.
2
2
also converges. By the Comparison Test we can
∞
∞
2n + 1
2n + 1
converges.
Finally,
the
series
converges.
4n
4n
n=3
73.
n=1
∞
∞ ln n
1
n2 −√3n
n=4
n
e
n=3
solution
By the comment preceding Exercise 31, we can choose N ≥ 4 so that for n ≥ N , ln n < n1/2 . Then
∞
n=N
∞
∞
ln n
n1/2
1
≤
=
2
2
3/2
n − 3n
n − 3n
n
− 3n1/2
n=N
To evaluate convergence of the latter series, let an =
Test:
n=N
1
1
and bn = 3/2 , and apply the Limit Comparison
n3/2 − 3n1/2
n
an
1
1
= lim 3/2
· n3/2 = lim
=0
1/2
n→∞ bn
n→∞ n
n→∞
− 3n
1 − 3n−1
L = lim
Thus
an converges if
bn does. But
bn is a convergent p-series. Thus
an converges and, by the comparison
test, so does the original series. Adding back in the finite number of terms for n < N does not affect convergence.
75.
∞
∞ 1
1
n1/2 lnlnnn
n=2
3
n=1
solution
By the comment preceding Exercise 31, we can choose N ≥ 2 so that for n ≥ N, ln n < n1/4 . Then
∞
n=N
∞
1
1
>
1/2
3/4
n ln n
n
n=N
which is a divergent p-series. Thus the original series diverges as well - as usual, adding back in the finite number of
terms for n < N does not affect convergence.
∞
2
∞ 4n + 15n
1
3n4 − 5n2 −417
n=1
3/2 − ln n
n
n=1
solution Apply the Limit Comparison Test with
77.
an =
May 23, 2011
4n2 + 15n
,
3n4 − 5n2 − 17
bn =
4n2
4
= 2
3n4
3n
666
C H A P T E R 10
INFINITE SERIES
We have
an
4n2 + 15n
12n4 + 45n3
12 + 45/n
3n2
= lim
=1
=
lim
·
= lim
n→∞ bn
n→∞ 3n4 − 5n2 − 17
n→∞ 12n4 − 20n2 − 68
n→∞ 12 − 20/n2 − 68/n4
4
L = lim
Now,
∞
n=1 bn is a p-series with p = 2 > 1, so converges. Since L = 1, we see that
∞
4n2 + 15n
n=1
3n4 − 5n2 − 17
converges as
well.
∞
1
∞
79. For
which a does
converge?
n
n(ln
n)a
4−n + 5−nn=2
n=1
First consider the case a > 0 but a = 1. Let f (x) =
solution
1
. This function is continuous, positive and
x(ln x)a
decreasing for x ≥ 2, so the Integral Test applies. Now,
∞
R
ln R
1
1
1
dx
dx
du
=
lim
=
lim
=
−
.
lim
a
1 − a R→∞ (ln R)a−1
R→∞ 2 x(ln x)a
R→∞ ln 2 ua
(ln 2)a−1
2 x(ln x)
Because
1
∞, 0 < a < 1
=
lim
R→∞ (ln R)a−1
0,
a>1
we conclude the integral diverges when 0 < a < 1 and converges when a > 1. Therefore
∞
n=2
1
converges for a > 1 and diverges for 0 < a < 1.
n(ln n)a
Next, consider the case a = 1. The series becomes
∞
n=2
1
1
. Let f (x) =
. For x ≥ 2, this function is continuous,
n ln n
x ln x
positive and decreasing, so the Integral Test applies. Using the substitution u = ln x, du = x1 dx, we find
R
ln R
∞
dx
dx
du
= lim
= lim
= lim (ln(ln R) − ln(ln 2)) = ∞.
u
R→∞ 2 x ln x
R→∞ ln 2
R→∞
2 x ln x
The integral diverges; hence, the series also diverges.
∞
(ln n)b
Finally, consider the case a < 0. Let b = −a > 0 so the series becomes
n=2
n
. Since ln n > 1 for all n ≥ 3, it
follows that
(ln n)b > 1
The series
∞
(ln n)b
n=2
n
∞
1
n=3
n
so
1
(ln n)b
> .
n
n
diverges, so by the Comparison Test we can conclude that
∞
(ln n)b
n=3
n
also diverges. Consequently,
diverges. Thus,
∞
n=2
1
diverges for a < 0.
n(ln n)a
To summarize:
∞
n=2
1
converges if a > 1 and diverges if a ≤ 1.
n(ln n)a
Approximating Infinite Sums
∞ In Exercises 81–83, let an = f (n), where f (x) is a continuous, decreasing function such
∞ 1
f (x) dx aconverges.
that f (x)
0 and a 1does
converge?
For≥which
n ln n
n=2
81. Show that
∞
∞
∞
f (x) dx ≤
an ≤ a1 +
f (x) dx
3
1
May 23, 2011
n=1
1
Convergence of Series with Positive Terms
S E C T I O N 10.3
solution
667
From the proof of the Integral Test, we know that
a2 + a3 + a4 + · · · + aN ≤
that is,
SN − a1 ≤
∞
N
1
f (x) dx ≤
or SN ≤ a1 +
f (x) dx
1
∞
f (x) dx;
1
∞
f (x) dx.
1
Also from the proof of the Integral test, we know that
N
f (x) dx ≤ a1 + a2 + a3 + · · · + aN−1 = SN − aN ≤ SN .
1
Thus,
N
1
f (x) dx ≤ SN ≤ a1 +
∞
f (x) dx.
1
Taking the limit as N → ∞ yields Eq. (3), as desired.
∞
Eq.
showasthat
83. Let S Using
=
an . (3),
Arguing
in Exercise 81, show that
n=1
M
an +
∞
n=1
M+1
∞
1
5≤
M+1
≤ 6 ∞
n1.2
f (x) dx ≤n=1
S≤
an +
f (x) dx
4
M+1
n=1
This series converges slowly. Use a computer algebra system to verify that SN < 5 for N ≤ 43,128 and S43,129 ≈
Conclude
that
5.00000021.
⎛
⎞
∞
M
0≤S−⎝
an +
f (x) dx ⎠ ≤ aM+1
5
M+1
n=1
This provides a method for approximating S with an error of at most aM+1 .
solution Following the proof of the Integral Test and the argument in Exercise 81, but starting with n = M + 1 rather
than n = 1, we obtain
∞
M+1
Adding
M
∞
f (x) dx ≤
an ≤ aM+1 +
∞
f (x) dx.
M+1
n=M+1
an to each part of this inequality yields
n=1
M
n=1
Subtracting
M
n=1
an +
∞
M+1
an +
∞
M+1
f (x) dx ≤
an = S ≤
n=1
M+1
an +
n=1
∞
f (x) dx.
M+1
f (x) dx from each part of this last inequality then gives us
⎛
0≤S−⎝
M
n=1
85.
∞
an +
∞
M+1
⎞
f (x) dx ⎠ ≤ aM+1 .
Apply Eq. (4) with M = 40,000 to show that
Use Eq. (4) with M = 43,129 to prove that
∞
∞ 1
1.644934066 ≤ 1≤ 1.644934068
n2
≤ 5.5915839
5.5915810 ≤
n=1 n1.2
n=1
Is this consistent with Euler’s result, according to which this infinite series has sum π 2 /6?
1
1
solution Using Eq. (4) with f (x) = 2 , an = 2 and M = 40,000, we find
x
n
∞
∞
∞
dx
1
dx
S40,000 +
≤
≤
S
+
.
40,001
n2
40,001 x 2
40,001 x 2
n=1
May 23, 2011
668
C H A P T E R 10
INFINITE SERIES
Now,
S40,000 = 1.6449090672;
S40,001 = S40,000 +
1
= 1.6449090678;
40,001
and
R
dx
dx
1
1
1
=
lim
=
−
lim
−
=
= 0.0000249994.
2
2
R
40,001
40,001
R→∞ 40,001 x
R→∞
40,001 x
∞
Thus,
1.6449090672 + 0.0000249994 ≤
∞
1
≤ 1.6449090678 + 0.0000249994,
n2
n=1
or
1.6449340665 ≤
∞
1
≤ 1.6449340672.
n2
n=1
π2
≈ 1.6449340668, our approximation is consistent with Euler’s result.
6
∞
87.
Using a CAS and Eq. (5), determine the value of∞ n−5 to within an error less than 10−4 .
Using a CAS and Eq. (5), determine
the value −5
of n=1
n−6 to within an error less than 10−4 . Check that your
−5
solution Using Eq. (5) with f (x) = x and an = n , we have
Since
n=1
⎛ proved that the sum is equal
⎞ to π 6 /945.
result is consistent with that of∞Euler, who
∞
M+1
0≤
n−5 − ⎝
n−5 +
x −5 dx ⎠ ≤ (M + 1)−5 .
n=1
n=1
M+1
To guarantee an error less than 10−4 , we need (M + 1)−5 ≤ 10−4 . This yields M ≥ 104/5 − 1 ≈ 5.3, so we choose
M = 6. Now,
7
n−5 = 1.0368498887,
n=1
and
∞
7
x −5 dx = lim
R
R→∞ 7
x −5 dx = −
1
1
lim R −4 − 7−4 =
= 0.0001041233.
4 R→∞
4 · 74
Thus,
∞
n=1
n−5 ≈
7
n=1
n−5 +
∞
7
x −5 dx = 1.0368498887 + 0.0001041233 = 1.0369540120.
∞
far canargument
a stack of
identical
books (of mass
and unit length)
without
tipping
over?
stackTest.
will
89. TheHow
following
proves
the divergence
of themharmonic
series Sextend
=
1/n without
using
theThe
Integral
not tip over if the (n + 1)st book is placed at the bottom of the stack with itsn=1
right edge located at the center of mass
Let of the first n books (Figure 5). Let cn be the center of mass of the first n books, measured along the x-axis, where we
take the positive x-axis to the left of the origin as in Figure 6. Recall that if an object of mass m1 has center of mass
1 1
1 1 1
at x1 and a second object of m
has1 +
center+of mass
+ · · ·x,2 , thenSthe
+ of+mass+of
· · the
· system has x-coordinate
S12 =
2 =center
3 5
2 4 6
m1 x1 + m2 x2
Show that if S converges, then
m1 + m2
(a) S1 and S2 also converge and S = S1 + S2 .
Show that if the1 (n + 1)st book is placed with its right edge at cn , then its center of mass is located at cn + 12 .
(b) S(a)
1 > S2 and S2 = 2 S.
(b) Consider
the first n books
as aconclude
single object
mass nm with center of mass at cn and the (n + 1)st book as a second
Observe
that (b) contradicts
(a), and
that Sofdiverges.
1
.
object ofAssume
mass m.throughout
Show that if
theS (n
+ 1)st book
placed
with
its right edge Write
at cn , then cn+1 = cn +
solution
that
converges;
we is
will
derive
a contradiction.
2(n + 1)
(c) Prove that lim cn = ∞. Thus, by using enough books, the stack can be extended as far as desired without
n→∞
1
1
1
an = , bn =
, cn =
tipping over.
n
2n − 1
2n
May 23, 2011
Convergence of Series with Positive Terms
S E C T I O N 10.3
669
for the nth terms in the series S, S1 , and S2 . Since 2n − 1 ≥ n for n ≥ 1, we have bn < an . Since S = an converges,
1
so does S1 =
bn by the Comparison Test. Also, cn = an , so again by the Comparison Test, the convergence of S
2
implies the convergence of S2 = cn . Now, define two sequences
b
n odd
bn = (n+1)/2
0
n even
0
n odd
cn =
cn/2 n even
and cn look
bn and cn
, but have
inserted in the “missing” places compared to an . Then an = bn + cn ;
That is, bn like
zeros
also S1 = bn = bn and S2 = cn = cn . Finally, since S, S1 , and S2 all converge, we have
S=
∞
an =
n=1
∞
∞
(bn + cn ) =
n=1
bn +
n=1
∞
cn =
n=1
∞
bn +
n=1
∞
cn = S1 + S2
n=1
1
Now, bn > cn for every n, so that S1 > S2 . Also, we showed above that cn = an , so that 2S2 = S. Putting all this
2
together gives
S = S1 + S2 > S2 + S2 = 2S2 = S
so that S > S, a contradiction. Thus S must diverge.
Further Insights and Challenges
∞
∞
91. Kummer’s Acceleration
Method Suppose
we
wish to approximate S =
1/n2 . There is a similar telescoping
−
ln
n
Let S =
an , where an = (ln(ln n))
.
n=1
n=2can be computed exactly (Example 1 in Section 10.2):
series whose value
n)) .
(a) Show, by taking logarithms, that an = n− ln(ln(ln
∞
12
ee . = 1
(b) Show that ln(ln(ln n)) ≥ 2 if n > C, where C =
n(ne + 1)
n=1
(c) Show that S converges.
(a) Verify that
S=
∞
n=1
∞ 1
+
n(n + 1)
n=1
1
1
−
n(n + 1)
n2
Thus for M large,
S ≈1+
M
n=1
1
n2 (n + 1)
6
(b) Explain what has been gained. Why is Eq. (6) a better approximation to S than is
M
1/n2 ?
n=1
Compute
(c)
1000
n=1
1
,
n2
1+
100
n=1
1
n2 (n + 1)
Which is a better approximation to S, whose exact value is π 2 /6?
solution
(a) Because the series
∞
∞
1
1
and
both converge,
n(n + 1)
n2
n=1
∞
n=1
1
+
n(n + 1)
n=1
∞ n=1
1
1
−
2
n(n + 1)
n
=
∞
n=1
∞
∞
∞
1
1
1
1
+
=
−
= S.
2
n(n + 1)
n(n + 1)
n
n2
n=1
n=1
Now,
n+1
1
1
n
1
= 2
−
−
= 2
,
n(n + 1)
n2
n (n + 1) n2 (n + 1)
n (n + 1)
May 23, 2011
n=1
670
C H A P T E R 10
INFINITE SERIES
so, for M large,
S ≈1+
M
n=1
(b) The series
∞
1
n=1 n2 (n+1) converges more rapidly than
1
.
n2 (n + 1)
∞
1
since the degree of n in the denominator is larger.
n2
n=1
(c) Using a computer algebra system, we find
1000
n=1
1
= 1.6439345667
n2
and
1+
100
n=1
1
= 1.6448848903.
n2 (n + 1)
The second sum is more accurate because it is closer to the exact solution
The series S =
∞
π2
≈ 1.6449340668.
6
k −3 has been computed to more than 100 million digits. The first 30 digits are
10.4 Absolute andk=1Conditional Convergence
S = 1.202056903159594285399738161511
S using
Acceleration
of Exercise
= 100 and auxiliary series
1. Approximate
Give an example
of a the
series
such that Method
an converges
but 91 with
|an |M
diverges.
(−1)n
1
∞
√
solution The series
Leibniz
Test, but the positive
series
is a divergent p-series.
√
−1
3 n converges byRthe
3
=
(n(n + 1)(n + 2)) .
n
2. Which of the following statements is equivalentn=1
to Theorem 1?
∞
∞
Exercise
46 in Section
R is a telescoping series with the sum R = 14 .
(a) According
If
|an | to
diverges,
then
an also10.2,
diverges.
Preliminary Questions
(b) If
(c) If
n=0
∞
n=0
∞
n=0
an diverges, then
∞
|an | also diverges.
n=0
∞
an converges, then
n=0
|an | also converges.
n=0
solution The correct answer is (b): If
∞
∞
an diverges, then
n=0
|an | also diverges. Take an = (−1)n n1 to see that
n=0
statements (a) and (c) are not true in general.
∞
√
(−1)n n is an alternating series and therefore converges. Is Lathika right?
3. Lathika argues that
n=1
solution
No. Although
∞
√
√
(−1)n n is an alternating series, the terms an = n do not form a decreasing sequence
n=1
that tends to zero. In fact, an =
∞
√
√
n is an increasing sequence that tends to ∞, so
(−1)n n diverges by the Divergence
n=1
Test.
4. Suppose that an is positive, decreasing, and tends to 0, and let S =
∞
(−1)n−1 an . What can we say about |S − S100 |
n=1
if a101 = 10−3 ? Is S larger or smaller than S100 ?
solution From the text, we know that |S − S100 | < a101 = 10−3 .Also, the Leibniz test tells us that S2N < S < S2N+1
for any N ≥ 1, so that S100 < S.
Exercises
1. Show that
∞
(−1)n
2n
n=0
converges absolutely.
solution The positive series
∞
1
1
is a geometric series with r = . Thus, the positive series converges, and the
2n
2
n=0
given series converges absolutely.
May 23, 2011
S E C T I O N 10.4
Absolute and Conditional Convergence
671
In Exercises
determine
whether
series converges
absolutely, conditionally, or not at all.
Show3–10,
that the
following
series the
converges
conditionally:
3.
∞
(−1)n−1
n1/3
∞
n=1
n=1
1
1
1
1
1
(−1)n−1 2/3 = 2/3 − 2/3 + 2/3 − 2/3 + · · ·
n
1
2
3
4
1 is positive, decreasing, and tends to zero; hence, the series
solution The sequence an = 1/3
n
by the Leibniz Test. However, the positive series
n=1
∞
n=1
∞
(−1)n−1
converges
n1/3
1
n1/3
is a divergent p-series, so the original series converges
conditionally.
5.
∞
(−1)n−1n 4
∞
(−1) n
(1.1)n
n=0
n3 + 1
n=1
solution The positive series
∞ 1 n
is a convergent geometric series; thus, the original series converges abso1.1
n=0
lutely.
7.
∞
(−1)n πn
∞
sin( )
n ln n 4
n=2
n2
n=1
1 . Then a forms a decreasing sequence (note that n and ln n are both increasing functions of
Let an = n ln
n
n
∞
∞
(−1)n
1
converges by the Leibniz Test. However, the positive series
n) that tends to zero; hence, the series
n ln n
n ln n
solution
n=2
n=2
diverges, so the original series converges conditionally.
∞
cos nπ n
∞
(−1)
(ln n)2 1
n=2
n=1 1 + n
solution Since cos nπ alternates between +1 and −1,
9.
∞
∞
cos nπ
(−1)n
=
(lnn)2
(lnn)2
n=2
n=2
This is an alternating series whose general term decreases to zero, so it converges. The associated positive series,
∞
n=2
1
(ln n)2
is a divergent series, so the original series converges conditionally.
∞
1
∞
11. Let S =cos n(−1)n+1 3 .
n
n=1
2n
n=1
(a) Calculate
Sn for 1 ≤ n ≤ 10.
(b) Use Eq. (2) to show that 0.9 ≤ S ≤ 0.902.
solution
(a)
S1 = 1
1
7
S2 = 1 − 3 = = 0.875
8
2
1
S3 = S2 + 3 = 0.912037037
3
1
S4 = S3 − 3 = 0.896412037
4
1
S5 = S4 + 3 = 0.904412037
5
May 23, 2011
1
S6 = S5 − 3 = 0.899782407
6
1
S7 = S6 + 3 = 0.902697859
7
1
S8 = S7 − 3 = 0.900744734
8
1
S9 = S8 + 3 = 0.902116476
9
1
S10 = S9 − 3 = 0.901116476
10
672
C H A P T E R 10
INFINITE SERIES
(b) By Eq. (2),
|S10 − S| ≤ a11 =
1
,
113
so
S10 −
1
1
≤ S ≤ S10 + 3 ,
113
11
or
0.900365161 ≤ S ≤ 0.901867791.
∞
(−1)n+1
Use Eq. (2) to approximate
13. Approximate
to three decimal places.
n4
n=1
∞
(−1)n+1
∞
n+1
1
n!
(−1)
n=1
, so that an = 4 . By
Eq. (2),
solution Let S =
n4
n
n=1
to four decimal places.
|SN − S| ≤ aN+1 =
1
.
(N + 1)4
To guarantee accuracy to three decimal places, we must choose N so that
1
< 5 × 10−4
(N + 1)4
or N >
√
4
2000 − 1 ≈ 5.7.
The smallest value that satisfies the required inequality is then N = 6. Thus,
1
1
1
1
1
S ≈ S6 = 1 − 4 + 4 − 4 + 4 − 4 = 0.946767824.
2
3
4
5
6
−5
In Exercises
Let15 and 16, find a value of N such that SN approximates the series with an error of at most 10 . If you have
a CAS, compute this value of SN .
∞
n
∞
(−1)n−1 2
S=
(−1)n+1
n
+1
15.
n=1
n(n + 2)(n + 3)
n=1
Use a computer algebra system to calculate and plot the partial sums Sn for 1 ≤ n ≤ 100. Observe that the partial
∞
n+1limit.
1
(−1)the
sums zigzag above
and below
solution Let S =
, so that an =
. By Eq. (2),
n (n + 2) (n + 3)
n (n + 2) (n + 3)
n=1
|SN − S| ≤ aN+1 =
1
.
(N + 1)(N + 3)(N + 4)
We must choose N so that
1
≤ 10−5
(N + 1)(N + 3)(N + 4)
or
(N + 1)(N + 3)(N + 4) ≥ 105 .
For N = 43, the product on the left hand side is 95,128, while for N = 44 the product is 101,520; hence, the smallest
value of N which satisfies the required inequality is N = 44. Thus,
S ≈ S44 =
44
n=1
(−1)n+1
= 0.0656746.
n(n + 2)(n + 3)
In Exercises
determine convergence or divergence by any method.
∞ 17–32,
(−1)n+1 ln n
∞
n!
n=1
7−n
17.
n=0
solution This is a (positive) geometric series with r =
May 23, 2011
1
< 1, so it converges.
7
S E C T I O N 10.4
19.
Absolute and Conditional Convergence
673
∞
∞ 1
1
5n − 3n
7.5
n=1
n
n=1
1
Use the Limit Comparison Test with n :
5
solution
1/(5n − 3n )
5n
1
=
lim
= lim
=1
n→∞
n→∞ 5n − 3n
n→∞ 1 − (3/5)n
1/5n
L = lim
But
∞
1
n=1 5n is a convergent geometric series. Since L = 1, the Limit Comparison Test tells us that the original series
converges as well.
21.
∞
∞ 1
n
3n4 +
12n
2
n=1
n −n
n=2
solution
1
:
3n4
Use the Limit Comparison Test with
L = lim
n→∞
(1/(3n4 + 12n)
3n4
1
=1
= lim
= lim
4
4
n→∞
n→∞
1 + 4n− 3
1/3n
3n + 12n
∞
1
1
= 13 ∞
n=1 n4 is a convergent p-series. Since L = 1, the Limit Comparison Test tells us that the original
3n4
series converges as well.
But
n=1
∞
∞ 1
(−1)n
2
n=1 n +21
n=1 n + 1
solution Apply the Limit Comparison Test and compare the series with the divergent harmonic series:
23.
√1
n
n2 +1
= lim = 1.
1
n→∞ n2 + 1
n
L = lim
n→∞
Because L > 0, we conclude that the series
∞
n=1
1
n2 + 1
diverges.
∞ n
n
3 + (−2)
∞
nn
(−1)
n
5
n=1
2
n=0 n + 1
solution The series
25.
∞ n
∞ n
3
3
=
5n
5
n=1
n=1
is a convergent geometric series, as is the series
∞
∞ 2 n
(−1)n 2n
=
.
−
5n
5
n=1
n=1
Hence,
∞ n
∞ n
∞ 3 + (−1)n 2n
3
2 n
=
+
−
5n
5
5
n=1
n=1
n=1
also converges.
27.
∞
3
∞ n 2 n+1
(−1)(−1)
n e−n /3
n=1
(2n + 1)!
n=1
solution
Consider the associated positive series
∞
3
n2 e−n /3 . This series can be seen to converge by the Integral
n=1
Test:
∞
1
3
x 2 e−x /3 dx = lim
R
R→∞ 1
3
3 R
3
x 2 e−x /3 dx = − lim e−x /3 1 = e−1/3 + lim e−R /3 = e−1/3 .
R→∞
The integral converges, so the original series converges absolutely.
May 23, 2011
R→∞
674
C H A P T E R 10
INFINITE SERIES
29.
∞
n
∞ (−1)
3
−n
1/2
n ne(ln n)/32
n=2
n=1
1
solution This is an alternating series with an = 1/2
. Because an is a decreasing sequence which converges
n (ln n)2
∞
(−1)n
to zero, the series
converges by the Leibniz Test. (Note that the series converges only conditionally, not
n1/2 (ln n)2
n=2
1
absolutely; the associated positive series is eventually greater than 3/4 , which is a divergent p-series).
n
∞
∞ln n
31.
1
n1.05
n=1
n(ln n)1/4
n=2
solution Choose N so that for n ≥ N we have ln n ≤ n0.01 . Then
∞
∞
∞
ln n
n0.01
1
≤
=
n1.05
n1.05
n1.04
n=N
n=N
n=N
This is a convergent p-series, so by the Comparison Test, the original series converges as well.
33. Show that
∞
1
1 1 1 1 1 1
S = − + − + − + ···
(ln n)2
2 2 3 3 4 4
n=2
converges by computing the partial sums. Does it converge absolutely?
solution The sequence of partial sums is
S1 =
1
2
1
=0
2
1
1
S3 = S2 + =
3
3
1
S4 = S3 − = 0
3
S2 = S1 −
and, in general,
⎧
⎨1,
SN = N
⎩
0,
for odd N
for even N
Thus, lim SN = 0, and the series converges to 0. The positive series is
N→∞
∞
1
1 1 1 1 1 1
+ + + + + + ··· = 2
;
2 2 3 3 4 4
n
n=2
which diverges. Therefore, the original series converges conditionally, not absolutely.
35.
Assumptions
Matter
Show by
The Leibniz
Test cannot
be applied
to counterexample that the Leibniz Test does not remain true if the sequence
an tends to zero but is not assumed nonincreasing. Hint: Consider
1
1
1
1
1 1
− 1+ · · 1· 1 1 1− 1+ 1 − 1 +
R = − +2 −3 +22 − 32 + 2· 3· · + 33 − n + · · ·
2 4 3 8 4 16
n 2
Why not? Show that it converges by another method.
solution Let
1
1
1
1 1 1 1 1
+ ··· +
− n+1 + · · ·
R= − + − + −
2 4 3 8 4 16
n+1 2
This is an alternating series with
an =
⎧
1
⎪
⎪
⎨ k + 1 , n = 2k − 1
⎪
⎪
⎩ 1 ,
2k+1
n = 2k
Note that an → 0 as n → ∞, but the sequence {an } is not decreasing. We will now establish that R diverges.
May 23, 2011
S E C T I O N 10.4
Absolute and Conditional Convergence
675
For sake of contradiction, suppose that R converges. The geometric series
∞
1
n=1
2n+1
converges, so the sum of R and this geometric series must also converge; however,
R+
∞
n=1
1
=
2n+1
∞
1
,
n
n=2
which diverges because the harmonic series diverges. Thus, the series R must diverge.
37. Prove
that if whether
an converges
absolutely,
an2conditionally:
also converges. Then give an example where
an is only
Determine
the
following
seriesthen
converges
2
conditionally convergent and
an diverges.
1 1 1 1 1 1 1 1
1
+ −
+ ···
1 − + − + − + − 3
2
5
3
7
4
9
5
11
solution Suppose the series
an converges absolutely. Because
|an | converges, we know that
lim |an | = 0.
n→∞
Therefore, there exists a positive integer N such that |an | < 1 for all n ≥ N. It then follows that for n ≥ N,
0 ≤ an2 = |an |2 = |an | · |an | < |an | · 1 = |an |.
By the Comparison Test we can then conclude that
an2 also converges.
∞
(−1)n
Consider the series
√ . This series converges by the Leibniz Test, but the corresponding positive series is a
n
n=1
∞
∞
∞
1
(−1)n
.
an2 is the divergent harmonic series
is conditionally convergent. Now,
√
n
n
n=1
n=1
n=1
Thus,
an2 need not converge if
an is only conditionally convergent.
divergent p-series; that is,
Further Insights and Challenges
39. Use Exercise 38 to show that the following series converges:
Prove the following variant of the Leibniz Test: If {an } is a positive, decreasing sequence with lim an = 0, then
n→∞
2
1
1
2
1
1
the series
−
+
+
−
+ ···
+
S=
ln 2 ln 3 ln 4 ln 5 ln 6 ln 7
a1 + a2 − 2a3 + a4 + a5 − 2a6 + · · ·
1
. Because an is
solution The given series has the structure of the generic series from Exercise 38 with an = ln(n+1)
is
increasing
and
bounded
by
a
+
a
,
and
continue
as
in
the
proof
the Leibniz
converges.
Hint:
Show
that
S
3N
1
2
a positive, decreasing sequence with lim an = 0, we can conclude from Exercise 38 that the given seriesofconverges.
n→∞
Test.
41. Show that the following series diverges:
Prove the conditional convergence of
1 1 2 1 1 1 2
1 +
3 +
1 +
1 −
1 +
3 ···
1 −
S =1+ +
R =1+
2 +
3 −
4 +
5 +
6 +
7 −
8 + ···
2 3 4 5 6 7 8
Hint: Use the result of Exercise 40 to write S as the sum of a convergent series and a divergent series.
solution
Let
R =1+
1 1 3 1 1 1 3
+ − + + + − + ···
2 3 4 5 6 7 8
S =1+
1 1 2 1 1 1 2
+ − + + + − + ···
2 3 4 5 6 7 8
and
For sake of contradiction, suppose the series S converges. From Exercise 40, we know that the series R converges. Thus,
the series S − R must converge; however,
S−R =
∞
1 1
1
11
+ +
+ ··· =
,
4 8 12
4
k
k=1
which diverges because the harmonic series diverges. Thus, the series S must diverge.
Prove that
∞
May 23, 2011
n=1
(−1)n+1
(ln n)a
n
a
676
C H A P T E R 10
INFINITE SERIES
43. We say that {bn } is a rearrangement of {an } if {bn } has the same terms as {an } but occurring in a different order. Show
∞
∞
an converges absolutely, then T =
bn also converges absolutely.
that if {bn } is a rearrangement of {an } and S =
n=1
n=1
(This result does not hold if S is only conditionally convergent.) Hint: Prove that the partial sums
|bn | are bounded.
n=1
It can be shown further that S = T .
solution
N
∞
Suppose the series S =
an converges absolutely and denote the corresponding positive series by
n=1
S+ =
∞
|an |.
n=1
Further, let TN =
N
|bn | denote the Nth partial sum of the series
n=1
∞
|bn |. Because {bn } is a rearrangement of {an }, we
n=1
know that
0 ≤ TN ≤
∞
|an | = S + ;
n=1
that is, the sequence {TN } is bounded. Moreover,
TN+1 =
N+1
|bn | = TN + |bN+1 | ≥ TN ;
n=1
that is, {TN } is increasing. It follows that {TN } converges, so the series
∞
|bn | converges, which means the series
n=1
∞
bn
n=1
converges absolutely.
Assumptions Matter In 1829, Lejeune Dirichlet pointed out that the great French mathematician Augustin
Louis Cauchy made a mistake in a published paper by improperly assuming the Limit Comparison Test to be valid
10.5
The Ratioseries.
andHere
Root
for nonpositive
are Tests
Dirichlet’s two series:
∞
∞
(−1)n
(−1)n
(−1)n
,
1
+
√
√
√
an+1 n
an n
n
or lim n=1
?
1. In the Ratio Test, is ρ equal to lim n=1
n→∞ an n→∞ an+1 Explain how they provide a counterexample to the Limit Comparison Test when the series are not assumed to be
positive. In the Ratio Test ρ is the limit lim an+1 .
solution
n→∞ an Preliminary Questions
2. Is the Ratio Test conclusive for
∞
∞
1
1
? Is it conclusive for
?
n
2
n
n=1
solution The general term of
∞
n=1
and
n=1
1
1
is an = n ; thus,
2n
2
an+1 1
2n
1
a = 2n+1 · 1 = 2 ,
n
a
1
ρ = lim n+1 = < 1.
n→∞ an
2
Consequently, the Ratio Test guarantees that the series
The general term of
∞
1
n=1
∞
1
converges.
2n
n=1
1
is an = ; thus,
n
n
May 23, 2011
an+1 n
n
1
a = n + 1 · 1 = n + 1,
n
S E C T I O N 10.5
and
The Ratio and Root Tests
677
a
n
= 1.
ρ = lim n+1 = lim
n→∞ an
n→∞ n + 1
The Ratio Test is therefore inconclusive for the series
∞
1
.
n
n=1
3. Can the Ratio Test be used to show convergence if the series is only conditionally convergent?
solution No. The Ratio Test can only establish absolute convergence and divergence, not conditional convergence.
Exercises
In Exercises 1–20, apply the Ratio Test to determine convergence or divergence, or state that the Ratio Test is inconclusive.
1.
∞
1
5n
n=1
solution With an = 51n ,
an+1 5n
1
1
a = 5n+1 · 1 = 5
n
Therefore, the series
an+1 1
= < 1.
and ρ = lim n→∞ an 5
∞
1
converges by the Ratio Test.
5n
n=1
∞
∞1
3.
(−1)n−1 n
nn
n=1
5n
n=1
solution With an = n1n ,
n
an+1 1
n
1 −n
1
nn
1
=
=
1
+
·
=
,
a (n + 1)n+1 1
n+1 n+1
n+1
n
n
and
a
1
ρ = lim n+1 = 0 · = 0 < 1.
n→∞ an
e
Therefore, the series
∞
1
converges by the Ratio Test.
nn
n=1
∞
∞ n
5.
3n + 2
n2 + 13
n=1
5n + 1
n=0
solution With an =
n ,
n2 +1
an+1 n+1
n2 + 1
n+1
n2 + 1
=
=
·
·
,
a (n + 1)2 + 1
n
n
n2 + 2n + 2
n
and
a
ρ = lim n+1 = 1 · 1 = 1.
n→∞ an
∞
n
, the Ratio Test is inconclusive.
n2 + 1
n=1
We can show that this series diverges by using the Limit Comparison Test and comparing with the divergent harmonic
series.
Therefore, for the series
∞
n
∞2 n
2
n100
n=1
n
n=1
n
solution With an = 2100 ,
7.
n
100
an+1 n100
n
2n+1
a = (n + 1)100 · 2n = 2 n + 1
n
Therefore, the series
∞
2n
diverges by the Ratio Test.
n100
n=1
May 23, 2011
and
an+1 = 2 · 1100 = 2 > 1.
ρ = lim n→∞ an 678
C H A P T E R 10
INFINITE SERIES
∞
10n 3
∞
n
n2
n=1 2 3n2
n=1
n
solution With an = 10n2 ,
9.
2
2
n+1
an+1 2n
1
= 10
·
= 10 · 2n+1
a 2
n
(n+1)
10
2
n
2
Therefore, the series
∞
10n
n2
n=1 2
and
a
ρ = lim n+1 = 10 · 0 = 0 < 1.
n→∞ an
converges by the Ratio Test.
∞ n
e n
∞
e
nn
n=1
n!
n=1
n
solution With an = ne n ,
n
an+1 en+1
nn
e
e
n
1 −n
=
·
=
=
,
1
+
a (n + 1)n+1 en
n+1 n+1
n+1
n
n
11.
and
a
1
ρ = lim n+1 = 0 · = 0 < 1.
n→∞ an
e
Therefore, the series
∞ n
e
converges by the Ratio Test.
nn
n=1
∞
n! 40
∞
13.
n
6n
n=0
n!
n=1
solution With an = 6n!n ,
an+1 (n + 1)! 6n
n+1
a = 6n+1 · n! = 6
n
Therefore, the series
an+1 = ∞ > 1.
and ρ = lim n→∞ an ∞
n!
diverges by the Ratio Test.
6n
n=0
∞
∞1
15.
n!
n ln n
n=2
n9
n=1
1 ,
solution With an = n ln
n
an+1 n ln n
n
ln n
1
a = (n + 1) ln(n + 1) · 1 = n + 1 ln(n + 1) ,
n
and
an+1 ln n
= 1 · lim
ρ = lim .
n→∞ an n→∞ ln(n + 1)
Now,
lim
ln n
n→∞ ln(n + 1)
= lim
ln x
x→∞ ln(x + 1)
= lim
x→∞
Thus, ρ = 1, and the Ratio Test is inconclusive for the series
Using the Integral Test, we can show that the series
∞
n=2
∞
n=2
1/(x + 1)
x
= lim
= 1.
x→∞ x + 1
1/x
1
.
n ln n
1
diverges.
n ln n
∞
2
∞ n
17.
1
(2n + 1)!
n=1
(2n)!
n=1
n2 ,
solution With an = (2n+1)!
2
2
an+1 1
= (n + 1) · (2n + 1)! = n + 1
,
a (2n + 3)!
n
(2n + 3)(2n + 2)
n2
n
May 23, 2011
S E C T I O N 10.5
The Ratio and Root Tests
and
a
ρ = lim n+1 = 12 · 0 = 0 < 1.
n→∞ a
n
Therefore, the series
∞
n=1
19.
n2
converges by the Ratio Test.
(2n + 1)!
∞
∞ 1
(n!)3
2n + 1
n=2
(3n)!
n=1
1
,
solution With an = n
2 +1
an+1 1 + 2−n
1
2n + 1
=
a = 2n+1 + 1 · 1
2 + 2−n
n
and
a
1
ρ = lim n+1 = < 1
n→∞ an
2
Therefore, the series
∞
n=2
1
converges by the Ratio Test.
2n + 1
∞
∞
nk 3−n converges for all exponents k.
21. Show
that1
ln nn=1
n=2
solution With an = nk 3−n ,
an+1 (n + 1)k 3−(n+1)
1
1 k
=
,
1
+
a =
3
n
nk 3−n
n
and, for all k,
a
1
1
ρ = lim n+1 = · 1 = < 1.
n→∞ an
3
3
Therefore, the series
∞
nk 3−n converges for all exponents k by the Ratio Test.
n=1
∞
∞n n
2 x 2 converges
if |x| < 1 .
23. Show that
n x n converges if |x| 2< 1.
Show that
solution
n=1
n=1
With an = 2n x n ,
an+1 2n+1 |x|n+1
= 2|x|
a = 2n |x|n
n
Therefore, ρ < 1 and the series
∞
a
and ρ = lim n+1 = 2|x|.
n→∞ an
2n x n converges by the Ratio Test provided |x| < 12 .
n=1
∞ n
r n
∞
25. Show that
rconverges if |r| < 1.
converges for all r.
Show that n
n=1
n!
n=1
n
solution With an = rn ,
an+1 |r|n+1
n
n
a = n + 1 · |r|n = |r| n + 1
n
Therefore, by the Ratio Test, the series
∞ n
r
converges provided |r| < 1.
n
n=1
May 23, 2011
a
and ρ = lim n+1 = 1 · |r| = |r|.
n→∞ an
679
680
C H A P T E R 10
INFINITE SERIES
∞
1 n
n!
∞ n
converges.
Hint:
Use
lim
= e.
1
+
27. Show that
2
n
n→∞
n
Is there any nvalue
of k such that
converges?
k
n=1
n
n=1
solution With an = nn!n ,
n n
an+1 1 −n
n
= (n + 1)! · n =
=
1
+
,
a (n + 1)n+1 n!
n+1
n
n
and
a
1
ρ = lim n+1 = < 1.
n→∞ an
e
Therefore, the series
∞
n!
converges by the Ratio Test.
nn
n=1
In Exercises 28–33, assume that |an+1 /an | converges to ρ = 13 . What can you say about the convergence of the given
series?
29.
∞
∞3
n an
nan
n=1
n=1
solution
Let bn = n3 an . Then
b
n + 1 3 an+1 1
3 1
ρ = lim n+1 = lim
a = 1 · 3 = 3 < 1.
n→∞ bn
n→∞
n
n
Therefore, the series
∞
n3 an converges by the Ratio Test.
n=1
31.
∞
∞n
3 ann
2 an
n=1
n=1
solution
Let bn = 3n an . Then
b
1
3n+1 an+1 = 3 · = 1.
ρ = lim n+1 = lim
n→∞ bn
n→∞ 3n an 3
Therefore, the Ratio Test is inconclusive for the series
∞
3n an .
n=1
33.
∞
∞2
an n
4 an
n=1
n=1
solution
Let bn = an2 . Then
2 2
b
a
1
1
ρ = lim n+1 = lim n+1 =
= < 1.
n→∞ bn
n→∞ an
3
9
Therefore, the series
∞
an2 converges by the Ratio Test.
n=1
∞
1 ∞ −1
Assume
an+1 /an converges
to ρ = 4. Does
35. Is the
Ratio that
Testconclusive
for the p-series
? n=1 an converge (assume that an = 0 for all n)?
np
n=1
solution With an = n1p ,
p
an+1 n
np
1
=
=
·
a (n + 1)p 1
n+1
n
Therefore, the Ratio Test is inconclusive for the p-series
∞
1
.
np
n=1
May 23, 2011
and
a
ρ = lim n+1 = 1p = 1.
n→∞ an
S E C T I O N 10.5
The Ratio and Root Tests
In Exercises 36–41, use the Root Test to determine convergence or divergence (or state that the test is inconclusive).
37.
∞
∞1
1
nn n
n=1
10
n=0
solution With an = n1n ,
√
n
Therefore, the series
an =
n
1
1
=
nn
n
and
lim
n→∞
√
n a = 0 < 1.
n
∞
1
converges by the Root Test.
nn
n=1
∞ ∞ k
k
k
39.
k
3k + 1
k=0
k + 10
k=0
k
k
solution With ak = 3k+1
,
√
k a =
k
Therefore, the series
∞ k=0
k
3k + 1
k
k
k
k
=
3k + 1
3k + 1
and
lim
√
k
k→∞
ak =
1
< 1.
3
k
converges by the Root Test.
−n2
∞ ∞ 1
1 −n
1+
1n+
n=4
n
n=1
−n2
solution With ak = 1 + n1
,
2 √
1 −n
1 −n
n
n a =
= 1+
1+
n
n
n
41.
Therefore, the series
∞ 1+
n=4
and
lim
n→∞
√
n a = e−1 < 1.
n
2
1 −n
converges by the Root Test.
n
In Exercises 43–56,∞determine
convergence or divergence using any method covered in the text so far.
2n2
2
diverges. Hint: Use 2n = (2n )n and n! ≤ nn .
that
∞ Prove
n!
2n + 4n
n=1
43.
7n
n=1
solution
Because the series
∞ n
∞ n
2
2
=
7n
7
n=1
and
n=1
∞ n
∞ n
4
4
=
7n
7
n=1
n=1
are both convergent geometric series, it follows that
∞ n
∞ n
∞ n
2
4
2 + 4n
=
+
7n
7
7
n=1
n=1
n=1
also converges.
45.
∞ 3
n 3
∞
n
5n
n=1
n!
n=1
3
solution The presence of the exponential term suggests applying the Ratio Test. With an = n5n ,
an+1 1 3
an+1 (n + 1)3 5n
1
1 3
=
= · 1 = 1 < 1.
·
=
and
ρ
=
lim
1
+
a n→∞ an 5
n
5
5
5n+1
n3
n
Therefore, the series
∞ 3
n
converges by the Ratio Test.
5n
n=1
May 23, 2011
681
682
C H A P T E R 10
INFINITE SERIES
∞
∞ 1
1
3 − n2
n=2 nn(ln
n)3
n=2
solution This series is similar to a p-series; because
47.
1
1
1
≈ √ = 3/2
3
n
n
n3 − n2
for large n, we will apply the Limit Comparison Test comparing with the p-series with p = 32 . Now,
√ 1
n3
n3 −n2
L = lim
=
lim
= 1.
1
n→∞
n→∞ n3 − n2
3/2
n
The p-series with p = 32 converges and L exists; therefore, the series
49.
∞
1
also converges.
3
2
n=2 n − n
∞
∞−0.82
n n + 4n
n=1
3n4 + 9
n=1
solution
∞
n−0.8 =
n=1
∞
1
n0.8
n=1
so that this is a divergent p-series.
51.
∞
∞−2n+1
4
(0.8)−n n−0.8
n=1
n=1
solution
Observe
∞
4−2n+1 =
n=1
∞
4 · (4−2 )n =
n=1
∞ n
1
4
16
n=1
1 ; therefore, this series converges.
is a geometric series with r = 16
∞
∞ 1 n−1
53.
sin (−1)
n2 √
n=1
n
n=1
solution Here, we will apply the Limit Comparison Test, comparing with the p-series with p = 2. Now,
L = lim
n→∞
sin 12
n
1
2
n
= lim
u→0
sin u
= 1,
u
where u = 12 . The p-series with p = 2 converges and L exists; therefore, the series
n
∞
n=1
1
sin 2 also converges.
n
∞
(−2)n
∞
√
1
n n cos
(−1)
n
solution Because
55.
n=1
n=1
√
2n
2x
2x ln 2
= lim 2x+1 x ln 2 = ∞ = 0,
lim √ = lim √ = lim
1
n→∞ n
x→∞ x
x→∞ √
x→∞
2 x
the general term in the series
∞
n=1
∞ (−2)n
√ does not tend toward zero; therefore, the series diverges by the Divergence Test.
n
n
n
Further
Insights
and Challenges
57.
n=1
∞
n + 12
√
Proof of the Root Test Let S =
an be a positive series, and assume that L = lim n an exists.
n→∞
n=0
n
(a) Show that S converges if L < 1. Hint:
Choose R with L < R < 1 and show that an ≤ R for n sufficiently large.
n
Then compare with the geometric series
R .
(b) Show that S diverges if L > 1.
May 23, 2011
S E C T I O N 10.5
solution
Suppose lim
n→∞
(a) If L < 1, let =
√
n
The Ratio and Root Tests
683
an = L exists.
1−L
. By the definition of a limit, there is a positive integer N such that
2
√
− ≤ n an − L ≤ for n ≥ N. From this, we conclude that
0≤
√
n a ≤ L+
n
for n ≥ N. Now, let R = L + . Then
R =L+
L+1
1+1
1−L
=
<
= 1,
2
2
2
and
0≤
for n ≥ N. Because 0 ≤ R < 1, the series
the Comparison Test. Therefore, the series
√
n
∞
an ≤ R
or
0 ≤ an ≤ R n
R n is a convergent geometric series, so the series
n=N
∞
∞
an converges by
n=N
an also converges.
n=0
(b) If L > 1, let =
L−1
. By the definition of a limit, there is a positive integer N such that
2
√
− ≤ n an − L ≤ for n ≥ N. From this, we conclude that
L− ≤
√
n a
n
for n ≥ N. Now, let R = L − . Then
R =L−
L+1
1+1
L−1
=
>
= 1,
2
2
2
and
R≤
for n ≥ N. Because R > 1, the series
Comparison Test. Therefore, the series
∞
√
n a
n
or
R n ≤ an
R n is a divergent geometric series, so the series
n=N
∞
∞
an diverges by the
n=N
an also diverges.
n=0
∞
cn n!
Ratio
Test
not apply, but verify convergence using the Comparison Test for the series
, where
c isdoes
a constant.
59. LetShow
S = that the
nn
n=1
1
1
1
1
1
+ 3 +if |c|
+
+ ···
(a) Prove that S converges absolutely if |c| < e +
and 2diverges
4 > 2e.5
2
3
2
3
n
√
e n!
(b) It is known that lim n+1/2 = 2π. Verify this numerically.
n→∞ n
(c) Use the Limit Comparison Test to prove that S diverges for c = e.
solution
n
(a) With an = cnnn! ,
n
an+1 |c|n+1 (n + 1)!
nn
1 −n
n
=
·
=
|c|
1
+
,
=
|c|
a |c|n n!
n+1
n
(n + 1)n+1
n
and
an+1 = |c|e−1 .
ρ = lim n→∞ an Thus, by the Ratio Test, the series
∞ n
c n!
converges when |c|e−1 < 1, or when |c| < e. The series diverges when
nn
n=1
|c| > e.
May 23, 2011
684
C H A P T E R 10
INFINITE SERIES
√
n n!
(b) The table below lists the value of en+1/2
for several increasing values of n. Since 2π = 2.506628275, the numerical
n
evidence verifies that
lim
en n!
n→∞ nn+1/2
=
√
2π.
n
100
1000
10000
100000
en n!
nn+1/2
2.508717995
2.506837169
2.506649163
2.506630363
(c) With c = e, the series S becomes
∞ n
e n!
. Using the result from part (b),
nn
n=1
en n!
nn
√
en n!
L = lim √ = lim n+1/2 = 2π.
n→∞
n→∞ n
n
Because the series
∞
∞ n
√
e n!
n diverges by the Divergence Test and L > 0, we conclude that
diverges by the Limit
nn
n=1
n=1
Comparison Test.
10.6 Power Series
Preliminary Questions
1. Suppose that
an x n converges for x = 5. Must it also converge for x = 4? What about x = −3?
solution The power series
an x n is centered at x = 0. Because the series converges for x = 5, the radius of
convergence must be at least 5 and the series converges absolutely at least for the interval |x| < 5. Both x = 4 and
x = −3 are inside this interval, so the series converges for x = 4 and for x = −3.
2. Suppose that
an (x − 6)n converges for x = 10. At which of the points (a)–(d) must it also converge?
(a) x = 8
(b) x = 11
(c) x = 3
(d) x = 0
solution The given power series is centered at x = 6. Because the series converges for x = 10, the radius of
convergence must be at least |10 − 6| = 4 and the series converges absolutely at least for the interval |x − 6| < 4, or
2 < x < 10.
(a) x = 8 is inside the interval 2 < x < 10, so the series converges for x = 8.
(b) x = 11 is not inside the interval 2 < x < 10, so the series may or may not converge for x = 11.
(c) x = 3 is inside the interval 2 < x < 10, so the series converges for x = 2.
(d) x = 0 is not inside the interval 2 < x < 10, so the series may or may not converge for x = 0.
3. What is the radius of convergence of F (3x) if F (x) is a power series with radius of convergence R = 12?
solution If the power series F (x) has radius of convergence R = 12, then the power series F (3x) has radius of
convergence R = 12
3 = 4.
4. The power series F (x) =
∞
nx n has radius of convergence R = 1. What is the power series expansion of F (x)
n=1
and what is its radius of convergence?
solution We obtain the power series expansion for F (x) by differentiating the power series expansion for F (x)
term-by-term. Thus,
F (x) =
∞
n2 x n−1 .
n=1
The radius of convergence for this series is R = 1, the same as the radius of convergence for the series expansion for
F (x).
May 23, 2011
S E C T I O N 10.6
Exercises
1. Use the Ratio Test to determine the radius of convergence R of
Power Series
685
∞ n
x
. Does it converge at the endpoints x = ±R?
2n
n=0
solution With an =
xn ,
2n
an+1 |x|n+1 2n
|x|
a = 2n+1 · |x|n = 2
n
a
|x|
and ρ = lim n+1 =
.
n→∞ an
2
|x|
By the Ratio Test, the series converges when ρ = |x|
2 < 1, or |x| < 2, and diverges when ρ =2 > 1, or |x| > 2.
n
The radius of convergence is therefore R = 2. For x = −2, the left endpoint, the series becomes ∞
n=0 (−1) , which is
∞
divergent. For x = 2, the right endpoint, the series becomes n=0 1, which is also divergent. Thus the series diverges at
both endpoints.
3. Show that the power series (a)–(c) have the same radius of convergence. Then show that (a) diverges at both endpoints,
∞
n
(b) converges at one endpoint but diverges
atxthe other, and (c) converges at both endpoints.
Use the Ratio Test to show that
√ n has radius of convergence R = 2. Then determine whether it converges
∞ n
∞
∞
n2
x
xn
xn
n=1
(b)
(c)
(a) at the nendpoints R = ±2.
3
n3n
n2 3n
n=1
n=1
n=1
solution
n
(a) With an = x3n ,
x x x n+1 3n an+1 = lim n+1 · n = lim = ρ = lim n→∞ an
n→∞ 3
x n→∞ 3
3
Then ρ < 1 if |x| < 3, so that the radius of convergence is R = 3. For the endpoint x = 3, the series becomes
∞ n
∞
3
=
1,
3n
n=1
n=1
which diverges by the Divergence Test. For the endpoint x = −3, the series becomes
∞
∞
(−3)n
=
(−1)n ,
3n
n=1
n=1
which also diverges by the Divergence Test.
n
x ,
(b) With an = n3
n
n+1
n
x an+1 x
n3
n
x
= .
= lim ·
=
lim
ρ = lim n
n→∞ an n→∞ (n + 1)3n+1
n→∞
x 3 n+1 3
Then ρ < 1 when |x| < 3, so that the radius of convergence is R = 3. For the endpoint x = 3, the series becomes
∞
∞
3n
1
,
=
n3n
n
n=1
n=1
which is the divergent harmonic series. For the endpoint x = −3, the series becomes
∞
∞
(−3)n
(−1)n
=
,
n3n
n
n=1
n=1
which converges by the Leibniz Test.
n
(c) With an = x2 n ,
n 3
2 x
an+1 n2 3n n
x n+1
x · n = lim = lim ρ = lim = n→∞ an n→∞ (n + 1)2 3n+1
n→∞
x
3 n+1 3
Then ρ < 1 when |x| < 3, so that the radius of convergence is R = 3. For the endpoint x = 3, the series becomes
∞
∞
3n
1
=
,
n2 3n
n2
n=1
May 23, 2011
n=1
686
C H A P T E R 10
INFINITE SERIES
which is a convergent p-series. For the endpoint x = −3, the series becomes
∞
∞
(−3)n
(−1)n
=
,
n2 3n
n2
n=1
n=1
which converges by the Leibniz Test.
∞
Repeat
for the following
5. Show
that Exercise
nn x n3diverges
for all x =series:
0.
∞
∞
n
(xn=0
− 5)
(x − 5)n
(b)
(a)
n
n
n
9 an = n x , and assuming x = 0, n9n
solution With
n=1
(c)
n=1
∞
(x − 5)n
n2 9n
n=1
n
(n + 1)n+1 x n+1 an+1 =∞
= lim x 1 + 1
ρ = lim =
lim
(n
+
1)
n
n
n→∞ an
n→∞ n→∞
n x
n
ρ < 1 only if x = 0, so that the radius of convergence is therefore R = 0. In other words, the power series converges
only for x = 0.
∞ 2n
√
∞ x
7. Use the Ratio Test to show that nn has radius of convergence R = 3.
3 converge?
n!x
For which values of x does
n=0
x 2n
n=0
solution With an = n ,
3
x 2(n+1) 3n x2 x2 an+1 =
lim
·
=
lim
ρ = lim = n→∞ an n→∞ 3n+1
x 2n n→∞ 3 3 Then ρ < 1 when |x 2 | < 3, or x =
√
√
3, so the radius of convergence is R = 3.
In Exercises 9–34, find
interval of convergence.
∞ the
x 3n+1
Show that
has radius of convergence R = 4.
∞
64n
n=0
nx n
9.
n=0
solution With an = nx n ,
(n + 1)x n+1 an+1 n + 1
= |x|
ρ = lim = lim = lim x
n→∞ an n→∞ n→∞
nx n
n Then ρ < 1 when |x| < 1, so that the radius of convergence is R = 1, and the series converges absolutely on the interval
∞
n, which diverges by the Divergence Test.
|x| < 1, or −1 < x < 1. For the endpoint x = 1, the series becomes
For the endpoint x = −1, the series becomes
∞
∞
n=0
(−1)n n, which also diverges by the Divergence Test. Thus, the series
n=1
nx n converges for −1 < x < 1 and diverges elsewhere.
n=0
11.
∞
x 2n+1
∞
(−1)2nn nn
x2 n
n=1
n
n=1
x 2n+1
,
2n n
x 2(n+1)+1
x2
n x 2 2n n ·
·
ρ = lim n+1
= = lim n→∞ 2
(n + 1) x 2n+1 n→∞ 2 n + 1 2 solution With an = (−1)n
√
√
Then ρ < 1 when |x| < 2, so the radius of convergence is R = 2, and the series converges absolutely on the interval
√
√
∞
∞
√
√
√
− 2
2
=
, which converges
(−1)n
(−1)n+1
− 2 < x < 2. For the endpoint x = − 2, the series becomes
n
n
n=1
n=1
√
∞
√
2
n
which also converges by the Leibniz test.
(−1)
by the Leibniz test. For the endpoint x = 2, the series becomes
n
Thus the series
∞
n=1
∞
n=0
n=1
√
√
x 2n+1
(−1)n n converges for − 2 ≤ x ≤ 2 and diverges elsewhere.
2 n
n
(−1)n n x 2n
4
May 23, 2011
S E C T I O N 10.6
13.
Power Series
687
∞ n
x
n5
n=4
n
solution With an = x 5 ,
n
5 x n+1
5
an+1 n
n
= lim ρ = lim ·
= lim x
= |x|
n→∞ an n→∞ (n + 1)5 x n n→∞ n+1 Then ρ < 1 when |x| < 1, so the radius of convergence is R = 1, and the series converges absolutely on the interval
∞
1
, which is a convergent p-series. For the
|x| < 1, or −1 < x < 1. For the endpoint x = 1, the series becomes
n5
n=1
∞
∞
(−1)n
,
which
converges
by
the
Leibniz
Test.
Thus,
the
series
endpoint x = −1, the series becomes
n5
n=1
n=4
for −1 ≤ x ≤ 1 and diverges elsewhere.
15.
xn
converges
n5
∞
n
∞x
(n!)n27 x n
n=0
n=8
n
solution With an = x 2 ,
(n!)
2 x n+1
an+1 1
(n!)2 · n = lim x
= lim ρ = lim =0
n→∞ an n→∞ ((n + 1)!)2
x n→∞ n+1 ρ < 1 for all x, so the radius of convergence is R = ∞, and the series converges absolutely for all x.
17.
∞
(2n)!n n
∞
8x
(n!)3 x n
n=0
n!
n=0
n
solution With an = (2n)!x3 , and assuming x = 0,
(n!)
(2(n + 1))!x n+1
a
(2n + 2)(2n + 1) (n!)3 x
·
=
lim
ρ = lim n+1 = lim n→∞ an
n→∞ ((n + 1)!)3
(2n)!x n n→∞ (n + 1)3
4n−1 + 6n−1 + 2n−3 4n2 + 6n + 2 = lim x 3
=0
= lim x
n→∞ n + 3n2 + 3n + 1 n→∞ 1 + 3n−1 + 3n−2 + n−3 Then ρ < 1 for all x, so the radius of convergence is R = ∞, and the series converges absolutely for all x.
19.
∞
(−1)n x nn
∞
4
2+1
x 2n−1
n=0 n(2n
+ 1)!
n=0
n n
√ x ,
solution With an = (−1)
2
n +1
(−1)n+1 x n+1
2 + 1
an+1 n
= lim ρ = lim ·
n→∞ an n→∞ n2 + 2n + 2 (−1)n x n 2+1
2
n
1
+
1/n
n2 + 1 = lim x
= lim x = lim x
2
2
n→∞ n + 2n + 2 n→∞ 1 + 2/n + 2/n n2 + 2n + 2 n→∞ = |x|
Then ρ < 1 when |x| < 1, so the radius of convergence is R = 1, and the series converges absolutely on the interval
∞
(−1)n
−1 < x < 1. For the endpoint x = 1, the series becomes
, which converges by the Leibniz Test. For the
2
n=1 n + 1
∞
1
, which diverges by the Limit Comparison Test comparing with the
endpoint x = −1, the series becomes
2
n=1 n + 1
∞
(−1)n x n
converges for −1 < x ≤ 1 and diverges elsewhere.
divergent harmonic series. Thus, the series
2
n=0 n + 1
∞
n=0
xn
n4 + 2
May 23, 2011
688
C H A P T E R 10
INFINITE SERIES
21.
∞
x 2n+1
3n + 1
n=15
solution With an =
x 2n+1
,
3n + 1
x 2n+3 3n + 1 an+1 2 3n + 1 = |x 2 |
ρ = lim ·
= lim = lim x
n→∞ an n→∞ 3n + 4 x 2n+1 n→∞
3n + 4 Then ρ < 1 when |x 2 | < 1, so the radius of convergence is R = 1, and the series converges absolutely for −1 < x < 1.
∞
1
, which diverges by the Limit Comparison Test comparing
For the endpoint x = 1, the series becomes
3n + 1
n=15
with the divergent harmonic series. For the endpoint x = −1, the series becomes
∞
n=15
−1
, which also diverges by
3n + 1
the Limit Comparison Test comparing with the divergent harmonic series. Thus, the series
∞
x 2n+1
converges for
3n + 1
n=15
−1 < x < 1 and diverges elsewhere.
∞
xn
∞
xn
ln n
n=2
n − 4 ln n
n=1
n
solution With an = lnx n ,
23.
x n+1
an+1 ln(n + 1) 1/(n + 1) ln
n
n = lim ρ = lim ·
=
lim
=
lim
= |x|
x
x
x
=
lim
n→∞ an n→∞ ln(n + 1) x n n→∞ ln n n→∞ 1/n n→∞ n + 1 using L’Hôpital’s rule. Then ρ < 1 when |x| < 1, so the radius of convergence is 1, and the series converges absolutely
∞
1
on the interval |x| < 1, or −1 < x < 1. For the endpoint x = 1, the series becomes
. Because ln1n > n1 and
ln n
n=2
∞
1
is the divergent harmonic series, the endpoint series diverges by the Comparison Test. For the endpoint x = −1,
n
n=2
the series becomes
∞
∞
(−1)n
xn
, which converges by the Leibniz Test. Thus, the series
converges for −1 ≤ x < 1
ln n
ln n
n=2
n=2
and diverges elsewhere.
∞
∞ 3n+2n
n(x x− 3)
n=1
ln n
n=2
solution With an = n(x − 3)n ,
25.
(n + 1)(x − 3)n+1 an+1 = lim (x − 3) · n + 1 = |x − 3|
ρ = lim =
lim
n
n→∞ an n→∞ n→∞
n(x − 3)
n Then ρ < 1 when |x − 3| < 1, so the radius of convergence is 1, and the series converges absolutely on the interval
∞
|x − 3| < 1, or 2 < x < 4. For the endpoint x = 4, the series becomes
n, which diverges by the Divergence Test.
For the endpoint x = 2, the series becomes
∞
∞
n=1
(−1)n n, which also diverges by the Divergence Test. Thus, the series
n=1
n(x − 3)n converges for 2 < x < 4 and diverges elsewhere.
n=1
∞
∞ n 5n
nn
(−1)(−5)
n (x(x−−7)3)
n=1
n2
n=1
solution With an = (−1)n n5 (x − 7)n ,
(−1)n+1 (n + 1)5 (x − 7)n+1 an+1 (n + 1)5 ρ = lim = lim = lim (x − 7) ·
n→∞ an n→∞ n→∞ (−1)n n5 (x − 7)n
n5 n5 + . . . = lim (x − 7) ·
= |x − 7|
n→∞ n5 27.
May 23, 2011
Power Series
S E C T I O N 10.6
689
Then ρ < 1 when |x − 7| < 1, so the radius of convergence is 1, and the series converges absolutely on the interval
∞
∞
(−1)2n n5 =
n5 , which diverges by the
|x − 7| < 1, or 6 < x < 8. For the endpoint x = 6, the series becomes
Divergence Test. For the endpoint x = 8, the series becomes
Thus, the series
∞
∞
n=1
n=1
(−1)n n5 , which also diverges by the Divergence Test.
n=1
(−1)n n5 (x − 7)n converges for 6 < x < 8 and diverges elsewhere.
n=1
29.
∞ n
2
∞
(x + 3)n
3n 27n (x − 1)3n+2
n=1
n=0
n
n
solution With an = 2 (x+3)
,
3n
2n+1 (x + 3)n+1
an+1 3n
2(x + 3) · 3n ·
=
lim
=
lim
ρ = lim n→∞ an n→∞ 3(n + 1)
2n (x + 3)n n→∞ 3n + 3 1
= |2(x + 3)|
= lim 2(x + 3) ·
n→∞
1 + 1/n Then ρ < 1 when |2(x + 3)| < 1, so when |x + 3| < 12 . Thus the radius of convergence is 12 , and the series converges
∞
1
absolutely on the interval |x + 3| < 12 , or − 72 < x < − 52 . For the endpoint x = − 52 , the series becomes
,
3n
n=1
which diverges because it is a multiple of the divergent harmonic series. For the endpoint x = − 72 , the series becomes
∞
∞ n
(−1)n
2
, which converges by the Leibniz Test. Thus, the series
(x + 3)n converges for − 72 ≤ x < − 52 and
3n
3n
n=1
n=1
diverges elsewhere.
31.
∞
(−5)n
∞
(x4)
+n10)n
(x −
n!
n=0
n!
n=0
n
n
solution With an = (−5)
n! (x + 10) ,
(−5)n+1 (x + 10)n+1
an+1 n!
5(x + 10) 1 = 0
ρ = lim ·
=
lim
=
lim
n
n
n→∞ an n→∞ n→∞
(n + 1)!
(−5) (x + 10)
n
Thus ρ < 1 for all x, so the radius of convergence is infinite, and
∞
(−5)n
(x + 10)n converges for all x.
n!
n=0
33.
∞
∞n
e (x − 2)n n
n! (x + 5)
n=12
n=10
solution With an = en (x − 2)n ,
en+1 (x − 2)n+1 an+1 ρ = lim = lim = lim |e(x − 2)| = |e(x − 2)|
n→∞ a n→∞ en (x − 2)n
n→∞
n
Thus ρ < 1 when |e(x − 2)| < 1, so when |x − 2| < e−1 . Thus the radius of convergence is e−1 , and the series converges
absolutely on the interval |x − 2| < e−1 , or 2 − e−1 < x < 2 + e−1 . For the endpoint x = 2 + e−1 , the series becomes
∞
∞
1, which diverges by the Divergence Test. For the endpoint x = 2 − e−1 , the series becomes
(−1)n , which also
n=1
diverges by the Divergence Test. Thus, the series
∞
n=12
elsewhere.
∞
(x + 4)n
(n ln n)2
n=2
May 23, 2011
n=1
en (x − 2)n converges for 2 − e−1 < x < 2 + e−1 and diverges
690
C H A P T E R 10
INFINITE SERIES
In Exercises 35–40, use Eq. (2) to expand the function in a power series with center c = 0 and determine the interval of
convergence.
1
1 − 3x
solution Substituting 3x for x in Eq. (2), we obtain
35. f (x) =
∞
∞
1
=
(3x)n =
3n x n .
1 − 3x
n=0
n=0
This series is valid for |3x| < 1, or |x| < 13 .
1
37. f (x) =
1
f (x) 3=− x
1 + 3x
solution First write
1
1
1
= ·
.
3−x
3 1 − x3
Substituting x3 for x in Eq. (2), we obtain
∞ ∞
x n xn
1
=
=
;
x
1− 3
3
3n
n=0
n=0
Thus,
∞
∞
xn
1 xn
1
=
=
.
n
n+1
3−x
3
3
3
n=0
n=0
This series is valid for |x/3| < 1, or |x| < 3.
1
39. f (x) =
1
f (x) 1=+ x 2
4 + 3x
solution Substituting −x 2 for x in Eq. (2), we obtain
∞
∞
1
2 )n =
=
(−x
(−1)n x 2n
1 + x2
n=0
n=0
This series is valid for |x| < 1.
41. Use the equalities
1
f (x) =
16 + 2x 3
− 13
1
1
=
=
1−x
−3 − (x − 4)
1 + x−4
3
to show that for |x − 4| < 3,
∞
(x − 4)n
1
=
(−1)n+1 n+1
1−x
3
n=0
solution
Substituting − x−4
3 for x in Eq. (2), we obtain
1
1 + x−4
3
=
∞ ∞
x−4 n (x − 4)n
=
(−1)n
.
−
3
3n
n=0
n=0
Thus,
∞
∞
n
(x − 4)n
1
1
n+1 (x − 4) .
=−
(−1)n
=
(−1)
1−x
3
3n
3n+1
n=0
n=0
This series is valid for | − x−4
3 | < 1, or |x − 4| < 3.
43. Use the method of Exercise 41 to expand 1/(4 − x) in a power series with center c = 5. Determine the interval of
Use the method of Exercise 41 to expand 1/(1 − x) in power series with centers c = 2 and c = −2. Determine
convergence.
the interval of convergence.
solution First write
1
1
1
=−
.
=
−1 − (x − 5)
1 + (x − 5)
4−x
May 23, 2011
S E C T I O N 10.6
Power Series
691
Substituting −(x − 5) for x in Eq. (2), we obtain
∞
∞
1
=
(−(x − 5))n =
(−1)n (x − 5)n .
1 + (x − 5)
n=0
n=0
Thus,
∞
∞
1
=−
(−1)n (x − 5)n =
(−1)n+1 (x − 5)n .
4−x
n=0
n=0
This series is valid for | − (x − 5)| < 1, or |x − 5| < 1.
45. Apply integration to the expansion
Find a power series that converges only for x in [2, 6).
∞
1
=
(−1)n x n = 1 − x + x 2 − x 3 + · · ·
1+x
n=0
to prove that for −1 < x < 1,
ln(1 + x) =
∞
x2
x3
x4
(−1)n−1 x n
=x−
+
−
+ ···
n
2
3
4
n=1
solution To obtain the first expansion, substitute −x for x in Eq. (2):
∞
∞
1
=
(−x)n =
(−1)n x n .
1+x
n=0
n=0
This expansion is valid for | − x| < 1, or −1 < x < 1.
Upon integrating both sides of the above equation, we find
ln(1 + x) =
dx
=
1+x
⎛
⎝
∞
⎞
(−1)n x n ⎠ dx.
n=0
Integrating the series term-by-term then yields
ln(1 + x) = C +
∞
n=0
(−1)n
x n+1
.
n+1
To determine the constant C, set x = 0. Then 0 = ln(1 + 0) = C. Finally,
ln(1 + x) =
∞
n=0
(−1)n
∞
x n+1
xn
=
(−1)n−1 .
n+1
n
n=1
47. Let F (x) = (x + 1) ln(1 + x) − x.
Use the result of Exercise 45 to prove that
(a) Apply integration to the result of Exercise 45 to prove that for −1 < x < 1,
1
1
1
1
3
∞
+
−
+ ···
ln = − n+1
2 n+1
2F (x)2 = 2 · 2(−1)
3 · 23x 4 · 24
n(n + 1)
Use your knowledge of alternating series to findn=1
an N such that the partial sum SN approximates ln 32 to within an
1 −3
. Confirm using a calculator to compute both SN and ln 32 .
error of atatmost
to prove
(b) Evaluate
x = 10
2
1
1
1
1
3 3 1
−
+
−
+ ···
ln − =
2 2 2
4 · 5 · 25
1 · 2 · 22
2 · 3 · 23
3 · 4 · 24
(c) Use a calculator to verify that the partial sum S4 approximates the left-hand side with an error no greater than the
term a5 of the series.
solution
(a) Note that
ln(x + 1) dx = (x + 1) ln(x + 1) − x + C
May 23, 2011
692
C H A P T E R 10
INFINITE SERIES
Then integrating both sides of the result of Exercise 45 gives
(x + 1) ln(x + 1) − x =
ln(x + 1) dx =
∞
(−1)n−1 x n
dx
n
n=1
For −1 < x < 1, which is the interval of convergence of the series in Exercise 45, therefore, we can integrate term by
term to get
(x + 1) ln(x + 1) − x =
∞
∞
∞
(−1)n−1
(−1)n−1 x n+1
x n+1
x n dx =
(−1)n+1
·
+C =
+C
n
n
n+1
n(n + 1)
n=1
n=1
n=1
(noting that (−1)n−1 = (−1)n+1 ). To determine C, evaluate both sides at x = 0 to get
0 = ln 1 − 0 = 0 + C
so that C = 0 and we get finally
(x + 1) ln(x + 1) − x =
∞
(−1)n+1
n=1
x n+1
n(n + 1)
(b) Evaluating the result of part(a) at x = 12 gives
∞
1
3 3 1
ln − =
(−1)n+1
2 2 2
n(n + 1)2n+1
n=1
1
1
1
1
=
−
+
−
+ ...
2
3
4
4 · 5 · 25
1·2·2
2·3·2
3·4·2
(c)
S4 =
1
1
1
1
−
+
−
= 0.1078125
4 · 5 · 25
1 · 2 · 22
2 · 3 · 23
3 · 4 · 24
1
≈ 0.0005208
a5 =
5 · 6 · 26
3 3 1
ln − ≈ 0.10819766
2 2 2
and
S4 − 3 ln 3 − 1 ≈ 0.0003852 < a5
2 2 2
49. Use the result of Example 7 to show that
Prove that for |x| < 1,
x2
x4
x6
x8
dx−
F (x) =
+ x 5 −x 9 + · · ·
+
=
x
−
1 · 42 3 · 4 5 · 6 7 ·−
8 ···
5
9
x +1
is an antiderivative of f (x) = tan−1 x satisfying
1/2 F (0) = 0. What is the radius of convergence of this power series?
Use the first two terms to approximate 0 dx/(x 4 + 1) numerically. Use the fact that you have an alternating series
solution
For the
−1 error
<x<
which
is the interval
convergence
to show that
in 1,
this
approximation
is atofmost
0.00022. for the power series for arctangent, we can integrate
term-by-term, so integrate that power series to get
F (x) =
tan−1 x dx =
∞ ∞
x 2n+2
(−1)n x 2n+1
dx =
(−1)n
2n + 1
(2n + 1)(2n + 2)
n=0
=
x2
1·2
−
x4
3·4
+
x6
5·6
n=0
−
x8
7·8
+ ··· + C
If we assume F (0) = 0, then we have C = 0. The radius of convergence of this power series is the same as that of the
original power series, which is 1.
∞
n
51. Evaluate
n . Hint: Use differentiation to show that
Verifyn=1
that2 function F (x) = x tan−1 x − 12 log(x 2 + 1) is an antiderivative of f (x) = tan−1 x satisfying
∞
F (0) = 0. Then use the result of Exercise 49
with
x = π6n−1
to show that
nx
(for |x| < 1)
(1 − x)−2 =
1 n=1 1
π
1
1
1 4
−
+
−
+ ···
√ − ln =
1 · 2(3) 3 · 4(32 ) 5 · 6(33 ) 7 · 8(34 )
6 3 2 3
Use a calculator to compare the value of the left-hand side with the partial sum S4 of the series on the right.
May 23, 2011
S E C T I O N 10.6
solution
Power Series
693
Differentiate both sides of Eq. (2) to obtain
∞
1
=
nx n−1 .
(1 − x)2
n=1
Setting x = 12 then yields
∞
n=1
n
= 2n−1
1
2 = 4.
1 − 12
Divide this equation by 2 to obtain
∞
n
= 2.
2n
n=1
53. Show that the following series converges absolutely for |x| < 1 and compute its sum:
Use the power series for (1 + x 2 )−1 and differentiation to prove that for |x| < 1,
F (x) = 1 − x − x 2 + x 3 − x 4 − x 5 + x 6 − x 7 − x 8 + · · ·
∞
2x
Hint: Write F (x) as a sum of three geometric series with
common
ratio(2n)x
x 3 . 2n−1
=
(−1)n−1
(x 2 + 1)2
solution Because the coefficients in the power series n=1
are all ±1, we find
a
r = lim n+1 = 1.
n→∞ an
The radius of convergence is therefore R = r −1 = 1, and the series converges absolutely for |x| < 1.
By Exercise 43 of Section 10.4, any rearrangement of the terms of an absolutely convergent series yields another
absolutely convergent series with the same sum as the original series. Following the hint, we now rearrange the terms of
F (x) as the sum of three geometric series:
F (x) = 1 + x 3 + x 6 + · · · − x + x 4 + x 7 + · · · − x 2 + x 5 + x 8 + · · ·
=
∞
(x 3 )n −
n=0
∞
n=0
x(x 3 )n −
∞
x 2 (x 3 )n =
n=0
1
x
x2
1 − x − x2
−
−
=
.
1 − x3
1 − x3
1 − x3
1 − x3
∞ n2
x
converges.
n!
n=1
1 + 2x
= 1 + x − 2x 2 + x 3 + x 4 − 2x 5 + x 6 + x 7 − 2x 8 + · · ·
n2
x
1
+
solution With an = n! , x + x 2
Show
that for
< 1,that
55. Find
all values
of |x|
x such
Hint: Use the hint from Exercise 53.
2
an+1 |x|(n+1)
n!
|x|2n+1
=
·
.
=
a 2
(n + 1)! |x|n
n+1
n
if |x| ≤ 1, then
|x|2n+1
= 0,
n→∞ n + 1
lim
and the series converges absolutely. On the other hand, if |x| > 1, then
|x|2n+1
= ∞,
n→∞ n + 1
lim
and the series diverges. Thus,
∞ n2
x
converges for −1 ≤ x ≤ 1 and diverges elsewhere.
n!
n=1
∞
n satisfying
Find
all values
such
series
converges: equation y = −y with initial condition y(0) = 1.
an xfollowing
the differential
57. Find
a power
seriesofPx(x)
= that the
n=0
F (x) = 1 + 3x + x 2 + 27x 3 + x 4 + 243x 5 + · · ·
Then use Theorem 1 of Section 5.8 to conclude that P (x) = e−x .
∞
solution Let P (x) =
an x n and note that P (0) = a0 ; thus, to satisfy the initial condition P (0) = 1, we must take
a0 = 1. Now,
n=0
P (x) =
∞
n=1
May 23, 2011
nan x n−1 ,
694
C H A P T E R 10
INFINITE SERIES
so
P (x) + P (x) =
∞
∞
nan x n−1 +
n=1
an x n =
n=0
∞
!
(n + 1)an+1 + an x n .
n=0
In order for this series to be equal to zero, the coefficient of x n must be equal to zero for each n; thus
(n + 1)an+1 + an = 0
or an+1 = −
an
.
n+1
Starting from a0 = 1, we then calculate
a
a1 = − 0 = −1;
1
1
a
a2 = − 1 = ;
2
2
a
1
1
a3 = − 2 = − = − ;
3
6
3!
and, in general,
an = (−1)n
1
.
n!
Hence,
∞
P (x) =
(−1)n
n=0
xn
.
n!
The solution to the initial value problem y = −y, y(0) = 1 is y = e−x . Because this solution is unique, it follows that
P (x) =
∞
(−1)n
n=0
xn
= e−x .
n!
59. Use the power series for y = ex to show that
x4
x6
x2
+
−
+ · · · .1
Let C(x) = 1 −
1
1
1
2!
4!
6!
=
− + − ···
2! 3! 4!
(a) Show that C(x) has an infinite radius ofe convergence.
−1 to=within
that C(x)
and f (x) series
= costo
x are
y partial
= −ysum
withSinitial
conditions y(0)
1, y (0)
= 0.
Use (b)
yourProve
knowledge
of alternating
findboth
an Nsolutions
such thatofthe
an error
N approximates e
−3
−1
This
initial
value
problem
has
a
unique
solution,
so
we
have
C(x)
=
cos
x
for
all
x.
of at most 10 . Confirm this using a calculator to compute both SN and e .
solution
Recall that the series for ex is
∞ n
x2
x3
x4
x
=1+x+
+
+
+ ··· .
n!
2!
3!
4!
n=0
Setting x = −1 yields
e−1 = 1 − 1 +
1
1
1
1
1
1
− + − +··· =
− + − +··· .
2! 3! 4!
2! 3! 4!
1 . The error in approximating e−1 with the partial sum S is therefore
This is an alternating series with an = (n+1)!
N
bounded by
|SN − e−1 | ≤ aN+1 =
1
.
(N + 2)!
To make the error at most 10−3 , we must choose N such that
1
≤ 10−3
(N + 2)!
or (N + 2)! ≥ 1000.
For N = 4, (N + 2)! = 6! = 720 < 1000, but for N = 5, (N + 2)! = 7! = 5040; hence, N = 5 is the smallest value
that satisfies the error bound. The corresponding approximation is
S5 =
1
1
1
1
1
− + − +
= 0.368055555
2! 3! 4! 5! 6!
Now, e−1 = 0.367879441, so
|S5 − e−1 | = 1.761 × 10−4 < 10−3 .
Let P (x) =
an x n be a power series solution to y = 2xy with initial condition y(0) = 1.
n=0
(a) ShowMay
that23,
the2011
odd coefficients a2k+1 are all zero.
(b) Prove that a2k = a2k−2 /k and use this result to determine the coefficients a2k .
S E C T I O N 10.6
Power Series
695
61. Find a power series P (x) satisfying the differential equation
y − xy + y = 0
9
with initial condition y(0) = 1, y (0) = 0. What is the radius of convergence of the power series?
∞
solution Let P (x) =
an x n . Then
n=0
P (x) =
∞
nan x n−1
and
P (x) =
n=1
∞
n(n − 1)an x n−2 .
n=2
Note that P (0) = a0 and P (0) = a1 ; in order to satisfy the initial conditions P (0) = 1, P (0) = 0, we must have a0 = 1
and a1 = 0. Now,
P (x) − xP (x) + P (x) =
∞
n(n − 1)an x n−2 −
n=2
=
∞
∞
nan x n +
n=1
(n + 2)(n + 1)an+2 x n −
n=0
∞
n=0
∞
nan x n +
n=1
= 2a2 + a0 +
∞
an x n
∞
an x n
n=0
!
(n + 2)(n + 1)an+2 − nan + an x n .
n=1
In order for this series to be equal to zero, the coefficient of x n must be equal to zero for each n; thus, 2a2 + a0 = 0 and
(n + 2)(n + 1)an+2 − (n − 1)an = 0, or
1
a2 = − a 0
2
and
an+2 =
n−1
an .
(n + 2)(n + 1)
Starting from a1 = 0, we calculate
a3 =
1−1
a1 = 0;
(3)(2)
a5 =
2
a3 = 0;
(5)(4)
a7 =
4
a5 = 0;
(7)(6)
and, in general, all of the odd coefficients are zero. As for the even coefficients, we have a0 = 1, a2 = − 12 ,
a4 =
1
1
a2 = − ;
(4)(3)
4!
a6 =
3
3
a4 = − ;
(6)(5)
6!
a8 =
5
15
a6 = −
(8)(7)
8!
and so on. Thus,
1
3
15
1
P (x) = 1 − x 2 − x 4 − x 6 − x 8 − · · ·
2
4!
6!
8!
To determine the radius of convergence, treat this as a series in the variable x 2 , and observe that
a2k+2 2k − 1
= lim
= 0.
r = lim k→∞ a2k k→∞ (2k + 2)(2k + 1)
Thus, the radius of convergence is R = r −1 = ∞.
63. Prove that
Find a power series satisfying Eq. (9) with initial condition y(0) = 0, y (0) = 1.
∞
(−1)k
x 2k+2
J2 (x) =
2k+2
2
k! (k + 3)!
k=0
is a solution of the Bessel differential equation of order 2:
x 2 y + xy + (x 2 − 4)y = 0
May 23, 2011
696
C H A P T E R 10
INFINITE SERIES
solution
Let J2 (x) =
∞
k=0
(−1)k
x 2k+2 . Then
22k+2 k! (k + 2)!
∞
(−1)k (k + 1)
k=0
22k+1 k! (k + 2)!
J2 (x) =
x 2k+1
∞
(−1)k (k + 1)(2k + 1) 2k
x
22k+1 k! (k + 2)!
J2 (x) =
k=0
and
x 2 J2 (x) + xJ2 (x) + (x 2 − 4)J2 (x) =
∞
∞
(−1)k (k + 1)(2k + 1) 2k+2 (−1)k (k + 1) 2k+2
+
x
x
22k+1 k! (k + 2)!
22k+1 k! (k + 2)!
k=0
−
=
(−1)k
k=0
22k+2 k! (k + 2)!
∞
k=0
=
∞
k=1
is it impossible
to expand f (x)
Further Why
Insights
and Challenges
Explain using Theorem 2.
65. Suppose that the coefficients of F (x) =
k=0
∞
x 2k+4 −
(−1)k k(k + 2) 2k+2
+
x
22k k!(k + 2)!
∞
(−1)k
k=0
22k k! (k + 2)!
x 2k+2
∞
(−1)k−1
k=1
22k (k − 1)! (k + 1)!
x 2k+2
∞
(−1)k
(−1)k
x 2k+2 −
x 2k+2 = 0.
2k
2k
2 (k − 1)!(k + 1)!
2 (k − 1)!(k + 1)!
k=1
= |x| as a power series that converges in an interval around x = 0?
∞
an x n are periodic; that is, for some whole number M > 0, we have
n=0
aM+n = an . Prove that F (x) converges absolutely for |x| < 1 and that
a + a1 x + · · · + aM−1 x M−1
F (x) = 0
1 − xM
Hint: Use the hint for Exercise 53.
solution Suppose the coefficients of F (x) are periodic, with aM+n = an for some whole number M and all n. The
F (x) can be written as the sum of M geometric series:
F (x) = a0 1 + x M + x 2M + · · · + a1 x + x M+1 + x 2M+1 + · · · +
= a2 x 2 + x M+2 + x 2M+2 + · · · + · · · + aM−1 x M−1 + x 2M−1 + x 3M−1 + · · ·
=
a0
a1 x
a2 x 2
aM−1 x M−1
a0 + a1 x + a2 x 2 + · · · + aM−1 x M−1
+
+
+
·
·
·
+
=
.
1 − xM
1 − xM
1 − xM
1 − xM
1 − xM
As each geometric series converges absolutely for |x| < 1, it follows that F (x) also converges absolutely for |x| < 1.
∞
Continuity of Power Series Let F (x) =
an x n be a power series with radius of convergence R > 0.
10.7
Taylor
Series
n=0
(a) Prove the inequality
Preliminary Questions
|x n − y n | ≤ n|x − y|(|x|n−1 + |y|n−1 )
1. Determine
f (0) and f (0) for a function f (x) with Maclaurin series
Hint: x n − y n = (x − y)(x n−1 + x n−2 y + · · · + y n−1 ).
·∞
T (x) = 3 + 2x + 12x 2 + 5x 3 + · · 2n|an |R1n converges. Hint: Show that
(b) Choose R1 with 0 < R1 < R. Show that the infinite series M =
n=0
solution The Maclaurin series for a function f has the form
n|an |R1n < |an |x n for all n sufficiently large if R1 < x < R.
(0)
(c) Use Eq. (10) to show that if |x| < Rf1 and
|y| <f R(0)
|Ff(x) (0)
− F (y)|
≤ M|x − y|.
1 , then
x+
x2 +
x3 + · · ·
f (0) +
1!
2! x. Hint: Choose
3!
(d) Prove that if |x| < R, then F (x) is continuous
at
R1 such that |x| < R1 < R. Show that if
> 0 is given, then |F (x) − F (y)| ≤ for all y such that |x − y| < δ, where
δ is any positive number that is less
f (0)
= 5. From this latter equation, it
Matching
this general
expression
with the
than /M
and R1 −
|x| (see Figure
6).given series, we find f (0) = 3 and
3!
follows that f (0) = 30.
May 23, 2011
S E C T I O N 10.7
Taylor Series
697
2. Determine f (−2) and f (4) (−2) for a function with Taylor series
T (x) = 3(x + 2) + (x + 2)2 − 4(x + 2)3 + 2(x + 2)4 + · · ·
solution The Taylor series for a function f centered at x = −2 has the form
f (−2) +
f (−2)
f (−2)
f (−2)
f (4) (−2)
(x + 2) +
(x + 2)2 +
(x + 2)3 +
(x + 2)4 + · · ·
1!
2!
3!
4!
Matching this general expression with the given series, we find f (−2) = 0 and
it follows that f (4) (−2) = 48.
f (4) (−2)
= 2. From this latter equation,
4!
3. What is the easiest way to find the Maclaurin series for the function f (x) = sin(x 2 )?
solution The easiest way to find the Maclaurin series for sin x 2 is to substitute x 2 for x in the Maclaurin series for
sin x.
4. Find the Taylor series for f (x) centered at c = 3 if f (3) = 4 and f (x) has a Taylor expansion
f (x) =
∞
(x − 3)n
n
n=1
Integrating the series for f (x) term-by-term gives
solution
∞
(x − 3)n+1
.
n(n + 1)
f (x) = C +
n=1
Substituting x = 3 then yields
f (3) = C = 4;
so
f (x) = 4 +
∞
(x − 3)n+1
.
n(n + 1)
n=1
5.
(a)
(b)
(c)
Let T (x) be the Maclaurin series of f (x). Which of the following guarantees that f (2) = T (2)?
T (x) converges for x = 2.
The remainder Rk (2) approaches a limit as k → ∞.
The remainder Rk (2) approaches zero as k → ∞.
solution The correct response is (c): f (2) = T (2) if and only if the remainder Rk (2) approaches zero as k → ∞.
Exercises
1. Write out the first four terms of the Maclaurin series of f (x) if
f (0) = 3,
f (0) = 2,
f (0) = 4,
f (0) = 12
solution The first four terms of the Maclaurin series of f (x) are
f (0) + f (0)x +
f (0) 2 f (0) 3
4
12
x +
x = 2 + 3x + x 2 + x 3 = 2 + 3x + 2x 2 + 2x 3 .
2!
3!
2
6
In Exercises
find
thefour
Maclaurin
and find
theof
interval
on whichatthe
expansion
is valid.
Write3–18,
out the
first
terms ofseries
the Taylor
series
f (x) centered
c=
3 if
3. f (x) =
solution
1
1 − 2x
f (3) = 1,
f (3) = 2,
f (3) = 12,
1 gives
Substituting 2x for x in the Maclaurin series for 1−x
∞
∞
1
(2x)n =
2n x n .
=
1 − 2x
n=0
This series is valid for |2x| < 1, or |x| < 12 .
May 23, 2011
n=0
f (3) = 3
698
C H A P T E R 10
INFINITE SERIES
5. f (x) = cos 3x x
f (x) =
solution Substituting
1 − x 4 3x for x in the Maclaurin series for cos x gives
cos 3x =
∞
(−1)n
n=0
∞
(3x)2n
9n x 2n
(−1)n
=
.
(2n)!
(2n)!
n=0
This series is valid for all x.
7. f (x) = sin(x 2 )
f (x) = sin(2x)
solution Substituting x 2 for x in the Maclaurin series for sin x gives
sin x 2 =
∞
(−1)n
n=0
∞
(x 2 )2n+1
x 4n+2
=
.
(−1)n
(2n + 1)!
(2n + 1)!
n=0
This series is valid for all x.
9. f (x) = ln(1 4x
− x2)
f (x) = e
solution Substituting −x 2 for x in the Maclaurin series for ln(1 + x) gives
ln(1 − x 2 ) =
∞
∞
∞ 2n
(−1)n−1 (−x 2 )n
(−1)2n−1 x 2n
x
=
=−
.
n
n
n
n=1
n=1
n=1
This series is valid for |x| < 1.
11. f (x) = tan−1 (x 2 ) −1/2
f (x) = (1 − x)
solution Substituting x 2 for x in the Maclaurin series for tan−1 x gives
tan−1 (x 2 ) =
∞
(−1)n
n=0
∞
(x 2 )2n+1
x 4n+2
=
.
(−1)n
2n + 1
2n + 1
n=0
This series is valid for |x| ≤ 1.
13. f (x) = ex−2 2
f (x) = x 2 ex
solution ex−2 = e−2 ex ; thus,
ex−2 = e−2
∞
∞ n
x
xn
=
.
n!
e2 n!
n=0
n=0
This series is valid for all x.
15. f (x) = ln(1 − 5x)
1 − cos x
f (x) Substituting
=
solution
−5x for x in the Maclaurin series for ln(1 + x) gives
x
ln(1 − 5x) =
∞
∞
∞ n n
(−1)n−1 (−5x)n
(−1)2n−1 5n x n
5 x
=
=−
.
n
n
n
n=1
n=1
n=1
This series is valid for |5x| < 1, or |x| < 15 , and for x = − 15 .
17. f (x) = sinh x
f (x) = (x 2 + 2x)ex
solution Recall that
sinh x =
1 x
(e − e−x ).
2
Therefore,
⎛
⎞
∞
∞
∞
1 ⎝ x n (−x)n ⎠ x n −
=
1 − (−1)n .
sinh x =
2
n!
n!
2(n!)
n=0
n=0
n=0
Now,
1 − (−1)n =
May 23, 2011
0, n even
2, n odd
S E C T I O N 10.7
Taylor Series
699
so
sinh x =
∞
2
k=0
∞
x 2k+1
x 2k+1
=
.
2(2k + 1)!
(2k + 1)!
k=0
This series is valid for all x.
In Exercises
f (x)19–28,
= coshfind
x the terms through degree four of the Maclaurin series of f (x). Use multiplication and substitution
as necessary.
19. f (x) = ex sin x
solution
Multiply the fourth-order Taylor Polynomials for ex and sin x:
x3
x4
x3
x2
+
+
x−
1+x+
2
6
24
6
= x + x2 −
x3
x3
x4
x4
+
−
+
+ higher-order terms
6
2
6
6
= x + x2 +
x3
+ higher-order terms.
3
The terms through degree four in the Maclaurin series for f (x) = ex sin x are therefore
x + x2 +
x3
.
3
sin x
21. f (x)
= = ex ln(1 − x)
f (x)
1−x
solution
Multiply the fourth order Taylor Polynomials for sin x and
x3
x−
6
1 + x + x2 + x3 + x4
1
:
1−x
= x + x2 −
x3
x4
+ x3 + x4 −
+ higher-order terms
6
6
= x + x2 +
5x 4
5x 3
+
+ higher-order terms.
6
6
The terms through order four of the Maclaurin series for f (x) =
x + x2 +
sin x
are therefore
1−x
5x 4
5x 3
+
.
6
6
23. f (x) = (1 + x)1/4
1
f (x) The
= first five generalized binomial coefficients for a = 1 are
solution
4
1 + sin x
1 −3
1 −3
−7
1 −3
−7
−11
1
3
7
−77
4
4
4
4
4
4
4
4
4
1,
,
=− ,
=
,
=
4
2!
32
3!
128
4!
2048
Therefore, the first four terms in the binomial series for (1 + x)1/4 are
3
1
7 3
77 4
x −
x
1 + x − x2 +
4
32
128
2048
25. f (x) = ex tan−1 x −3/2
f (x) = (1 + x)
solution Using the Maclaurin series for ex and tan−1 x, we find
x3
x3
x3
x4
x4
x2
x3
x
−1
+
+ ···
x−
+ · · · = x + x2 −
+
+
−
+ ···
e tan x = 1 + x +
2
6
3
3
2
6
3
1
1
= x + x2 + x3 − x4 + · · · .
6
6
May 23, 2011
700
C H A P T E R 10
INFINITE SERIES
27. f (x) = esin x
f (x) = sin (x 3 − x)
solution Substituting sin x for x in the Maclaurin series for ex and then using the Maclaurin series for sin x, we find
sin3 x
sin4 x
sin2 x
+
+
+ ···
2
6
24
2
1
x3
1
1
x3
+ ··· +
x−
+ ···
+ (x − · · · )3 +
=1+ x−
(x − · · · )4
6
2
6
6
24
esin x = 1 + sin x +
1
1
1
1
1
= 1 + x + x2 − x3 + x3 − x4 + x4 + · · ·
2
6
6
6
24
1
1
= 1 + x + x2 − x4 + · · · .
2
8
In Exercises 29–38, find the Taylor series centered at c and find the interval on which the expansion is valid.
x
f (x) 1= e(e )
29. f (x) = , c = 1
x
solution Write
1
1
=
,
x
1 + (x − 1)
1 to obtain
and then substitute −(x − 1) for x in the Maclaurin series for 1−x
∞
∞
1
[−(x − 1)]n =
=
(−1)n (x − 1)n .
x
n=0
n=0
This series is valid for |x − 1| < 1.
1
31. f (x)
= = e3x,, cc =
= 5−1
f (x)
1−x
solution Write
1
1
1
1
.
=
=− ·
1−x
−4 − (x − 5)
4 1 + x−5
4
1
Substituting − x−5
4 for x in the Maclaurin series for 1−x yields
∞ ∞
n
x−5 n n (x − 5) .
=
=
(−1)
−
n
4
4
1 + x−5
4
n=0
n=0
1
Thus,
∞
∞
n
1
(x − 5)n
1
n+1 (x − 5) .
=−
(−1)n
=
(−1)
1−x
4
4n
4n+1
n=0
n=0
This series is valid for x−5
4 < 1, or |x − 5| < 4.
33. f (x) = x 4 + 3x − 1, cπ= 2
f (x) = sin x, c =
solution To determine the2 Taylor series with center c = 2, we compute
f (x) = 4x 3 + 3,
f (x) = 12x 2 ,
f (x) = 24x,
and f (4) (x) = 24. All derivatives of order five and higher are zero. Now,
f (2) = 21,
f (2) = 35,
f (2) = 48,
f (2) = 48,
and f (4) (2) = 24. Therefore, the Taylor series is
21 + 35(x − 2) +
48
48
24
(x − 2)2 + (x − 2)3 + (x − 2)4 ,
2
6
24
or
21 + 35(x − 2) + 24(x − 2)2 + 8(x − 2)3 + (x − 2)4 .
May 23, 2011
Taylor Series
S E C T I O N 10.7
1
35. f (x)
= =2 ,x 4 c+=3x4− 1, c = 0
f (x)
x
solution We will first find the Taylor series for x1 and then differentiate to obtain the series for 12 . Write
x
1
1
1
1
=
= ·
.
x
4 + (x − 4)
4 1 + x−4
4
1
Now substitute − x−4
4 for x in the Maclaurin series for 1−x to obtain
∞ ∞
1
x−4 n (x − 4)n
1
=
(−1)n n+1 .
−
=
x
4 n=
4
4
n=0
Differentiating term-by-term yields
∞
1
(x − 4)n−1
− 2 =
(−1)n n
,
x
4n+1
n=1
so that
∞
∞
1
(x − 4)n−1
(x − 4)n
=
(−1)n−1 n
=
(−1)n (n + 1) n+2 .
2
n+1
x
4
4
n=1
n=0
This series is valid for x−4
4 < 1, or |x − 4| < 4.
1√
37. f (x)
= = x,
, cc==43
f (x)
1 − x2
solution
By partial fraction decomposition
1
1
1
= 2 + 2 ,
1−x
1+x
1 − x2
so
1
1
1
1
1
1
1
2
2
+
=− ·
=
+ ·
.
x−3
2
−2
−
(x
−
3)
4
+
(x
−
3)
4
8
1−x
1+ 2
1 + x−3
4
1
Substituting − x−3
2 for x in the Maclaurin series for 1−x gives
∞ ∞
x − 3 n (−1)n
=
=
(x − 3)n ,
−
2
2n
1 + x−3
2
n=0
n=0
1
while substituting − x−3
4 for x in the same series gives
∞ ∞
x − 3 n (−1)n
=
=
(x − 3)n .
−
x−3
4
4n
1+ 4
n=0
n=0
1
Thus,
∞
∞
∞
∞
1
1 (−1)n
(−1)n
(−1)n+1
(−1)n
n+ 1
n=
n+
=
−
(x
−
3)
(x
−
3)
(x
−
3)
(x − 3)n
n
n
2
n+2
4
2
8
4
1−x
2
22n+3
n=0
n=0
n=0
n=0
∞
∞
(−1)n+1
(−1)n+1 (2n+1 − 1)
(−1)n
=
+ 2n+3 (x − 3)n =
(x − 3)n .
n+2
2
2
22n+3
n=0
n=0
This series is valid for |x − 3| < 2.
39. Use the identity1 cos2 x = 12 (1 + cos 2x) to find the Maclaurin series for cos2 x.
, c = −1
f (x) =
−2
solution The3xMaclaurin
series for cos 2x is
∞
n=0
May 23, 2011
(−1)n
∞
(2x)2n
22n x 2n
=
(−1)n
(2n)!
(2n)!
n=0
701
702
C H A P T E R 10
INFINITE SERIES
so the Maclaurin series for cos2 x = 12 (1 + cos 2x) is
n 22n x 2n
1+ 1+ ∞
n=1 (−1) (2n)!
2
=1+
∞
(−1)n
n=1
22n−1 x 2n
(2n)!
41. Use the Maclaurin series for ln(1 + x) and ln(1 − x) to show that
Show that for |x| < 1,
1+x
x33
x55
1
ln
=
x
+
+
+ ···
x
x
2 tanh
1 −−1x x = x + 3 + 5 + · · ·
3
5
for |x| < 1. What can you conclude by comparing this result with that of Exercise 40?
1
d
tanh−1 xseries
= for 2ln. (1 + x) and ln (1 − x), we have for |x| < 1
Hint: Recall
thatthe Maclaurin
solution
Using
dx
1−x
ln(1 + x) − ln(1 − x) =
∞
∞
(−1)n−1 n (−1)n−1
x −
(−x)n
n
n
n=1
=
n=1
∞
∞
∞
(−1)n−1 n x n
1 + (−1)n−1 n
x +
=
x .
n
n
n
n=1
n=1
n=1
Since 1 + (−1)n−1 = 0 for even n and 1 + (−1)n−1 = 2 for odd n,
ln (1 + x) − ln (1 − x) =
∞
k=0
2
x 2k+1 .
2k + 1
Thus,
∞
∞
1
x 2k+1
1
1 2
1+x
ln
= (ln(1 + x) − ln(1 − x)) =
x 2k+1 =
.
2
1−x
2
2
2k + 1
2k + 1
k=0
k=0
Observe that this is the same series we found in Exercise 40; therefore,
1+x
1
ln
= tanh−1 x.
2
1−x
1
, that for |x| < 1,
43. Show, by integrating the Maclaurin series for
1 f (x) = 1
2 Maclaurin series of
Differentiate the Maclaurin series for
twice to 1find
.
− xthe
1−x
(1 − x)3
∞
1 · 3 · 5 · · · (2n − 1) x 2n+1
sin−1 x = x +
2 · 4 · 6 · · · (2n) 2n + 1
n=1
solution
From Example 10, we know that for |x| < 1
∞
∞
1 · 3 · 5 · · · (2n − 1) 2n
1 · 3 · 5 · · · (2n − 1) 2n
1
x =1+
x ,
=
2
2
·
4
·
6
·
·
·
(2n)
2 · 4 · 6 · · · (2n)
1−x
n=0
n=1
so, for |x| < 1,
sin−1 x =
∞
1 · 3 · 5 · · · (2n − 1) x 2n+1
dx
.
=C+x+
2 · 4 · 6 · · · (2n) 2n + 1
1 − x2
n=1
Since sin−1 0 = 0, we find that C = 0. Thus,
sin−1 x = x +
∞
1 · 3 · 5 · · · (2n − 1) x 2n+1
.
2 · 4 · 6 · · · (2n) 2n + 1
n=1
45. How many terms of the Maclaurin series of f (x) = ln(1 + x) are needed to compute ln 1.2 to within an error of at
−1 1
Use the
first the
fivecomputation
terms of theand
Maclaurin
in Exercise
43calculator
to approximate
most 0.0001?
Make
compareseries
the result
with the
value. sin 2 . Compare the result with
the calculator value.
solution Substitute x = 0.2 into the Maclaurin series for ln (1 + x) to obtain:
ln 1.2 =
∞
n=1
May 23, 2011
(−1)n−1
∞
(0.2)n
1
=
(−1)n−1 n .
n
5 n
n=1
S E C T I O N 10.7
This is an alternating series with an =
Taylor Series
703
1
. Using the error bound for alternating series
n · 5n
|ln 1.2 − SN | ≤ aN+1 =
1
,
(N + 1)5N+1
so we must choose N so that
1
< 0.0001
(N + 1)5N+1
or (N + 1)5N+1 > 10,000.
For N = 3, (N + 1)5N+1 = 4 · 54 = 2500 < 10, 000, and for N = 4, (N + 1)5N+1 = 5 · 55 = 15, 625 > 10, 000;
thus, the smallest acceptable value for N is N = 4. The corresponding approximation is:
S4 =
4
(−1)n−1
n=1
=
5n · n
1
1
1
1
− 2
+ 3
− 4
= 0.182266666.
5 5 ·2 5 ·3 5 ·4
Now, ln 1.2 = 0.182321556, so
|ln 1.2 − S4 | = 5.489 × 10−5 < 0.0001.
2
2
47. UseShow
the Maclaurin
expansion for e−t to express the function F (x) = 0x e−t dt as an alternating power series in x
that
(Figure 4).
5
π 3 to πapproximate
π7
(a) How many terms of the Maclaurin series areπ needed
−
+
−
+ · the
· · integral for x = 1 to within an error of at
3!
5!
7!
most 0.001?
Carry
out the
computation
check
your answer
using
a computer
algebra system.
(b) converges
to zero.
How
many termsand
must
be computed
to get
within
0.01 of zero?
y
F(x)
T15(x)
x
1
2
FIGURE 4 The Maclaurin polynomial T15 (x) for F (t) =
solution
x
e−t dt.
2
0
Substituting −t 2 for t in the Maclaurin series for et yields
e−t =
2
∞
∞
(−t 2 )n
t 2n
=
;
(−1)n
n!
n!
n=0
n=0
thus,
x
0
e−t dt =
2
∞
(−1)n
x 2n+1
.
n!(2n + 1)
(−1)n
1
.
n!(2n + 1)
n=0
(a) For x = 1,
1
0
e−t dt =
2
∞
n=0
1
; therefore, the error incurred by using SN to approximate the value of
This is an alternating series with an = n!(2n+1)
the definite integral is bounded by
1
2
1
−t
.
e
dt − SN ≤ aN+1 =
0
(N + 1)!(2N + 3)
To guarantee the error is at most 0.001, we must choose N so that
1
< 0.001
(N + 1)!(2N + 3)
or (N + 1)!(2N + 3) > 1000.
For N = 3, (N + 1)!(2N + 3) = 4! · 9 = 216 < 1000 and for N = 4, (N + 1)!(2N + 3) = 5! · 11 = 1320 > 1000;
thus, the smallest acceptable value for N is N = 4. The corresponding approximation is
S4 =
4
n=0
May 23, 2011
1
1
1
1
(−1)n
=1− +
−
+
= 0.747486772.
n!(2n + 1)
3 2! · 5 3! · 7 4! · 9
704
C H A P T E R 10
INFINITE SERIES
(b) Using a computer algebra system, we find
1
0
therefore
e−t dt = 0.746824133;
2
1
2
−t
e
dt − S4 = 6.626 × 10−4 < 10−3 .
0
x
−4
In Exercises 49–52, express
sin tthe
dt definite integral as an infinite series and find its value to within an error of at most 10 .
Let F (x) =
. Show that
1
t
0
cos(x 2 ) dx
49.
0
x3
x5
x7
F (x) = x −
+
−
+ ···
2
3 · 3!
5 ·x5!yields
7 · 7!
solution Substituting x for x in the Maclaurin series
for cos
Evaluate F (1) to three decimal places. ∞
∞
(x 2 )2n
x 4n
=
;
(−1)n
(−1)n
cos(x 2 ) =
(2n)!
(2n)!
n=0
n=0
therefore,
1
0
cos(x 2 ) dx =
∞
(−1)n
n=0
1
∞
x 4n+1
(−1)n
.
=
(2n)!(4n + 1) (2n)!(4n + 1)
0
n=0
1
; therefore, the error incurred by using SN to approximate the value of
This is an alternating series with an = (2n)!(4n+1)
the definite integral is bounded by
1
1
2
.
cos(x ) dx − SN ≤ aN+1 =
0
(2N + 2)!(4N + 5)
To guarantee the error is at most 0.0001, we must choose N so that
1
< 0.0001
(2N + 2)!(4N + 5)
or (2N + 2)!(4N + 5) > 10,000.
For N = 2, (2N + 2)!(4N + 5) = 6! · 13 = 9360 < 10,000 and for N = 3, (2N + 2)!(4N + 5) = 8! · 17 = 685,440 >
10,000; thus, the smallest acceptable value for N is N = 3. The corresponding approximation is
S3 =
n=0
1
3
−x
e 1 dx
51.
tan−1 (x 2 ) dx
0
0
solution
3
(−1)n
1
1
1
=1−
+
−
= 0.904522792.
(2n)!(4n + 1)
5 · 2! 9 · 4! 13 · 6!
Substituting −x 3 for x in the Maclaurin series for ex yields
e−x =
3
∞
∞
(−x 3 )n
x 3n
=
;
(−1)n
n!
n!
n=0
n=0
therefore,
1
0
3
e−x dx =
∞
(−1)n
n=0
1
∞
x 3n+1 (−1)n
.
=
n!(3n + 1) n!(3n + 1)
0
n=0
1
; therefore, the error incurred by using SN to approximate the value of
This is an alternating series with an = n!(3n+1)
the definite integral is bounded by
1
3
1
−x
.
e
dx − SN ≤ aN+1 =
0
(N + 1)!(3N + 4)
To guarantee the error is at most 0.0001, we must choose N so that
1
< 0.0001
(N + 1)!(3N + 4)
May 23, 2011
or (N + 1)!(3N + 4) > 10,000.
S E C T I O N 10.7
Taylor Series
705
For N = 4, (N + 1)!(3N + 4) = 5! · 16 = 1920 < 10,000 and for N = 5, (N + 1)!(3N + 4) = 6! · 19 = 13,680 >
10,000; thus, the smallest acceptable value for N is N = 5. The corresponding approximation is
S5 =
5
n=0
(−1)n
= 0.807446200.
n!(3n + 1)
1 53–56, express the integral as an infinite series.
In Exercises
dx
x
10 − cos(t)
x 4 + 1dt, for all x
53.
t
0
solution The Maclaurin series for cos t is
cos t =
∞
(−1)n
n=0
∞
t 2n
t 2n
=1+
,
(−1)n
(2n)!
(2n)!
n=1
so
1 − cos t = −
∞
∞
t 2n
t 2n
=
,
(−1)n+1
(2n)!
(2n)!
(−1)n
n=1
n=1
and
∞
∞
1
t 2n
t 2n−1
1 − cos t
=
(−1)n+1
(−1)n+1
=
.
t
t
(2n)!
(2n)!
n=1
n=1
Thus,
x
x
∞
∞
1 − cos(t)
t 2n x 2n
n+1
(−1)
(−1)n+1
dt =
.
=
t
(2n)!2n (2n)!2n
0
n=1
0
n=1
x
x
2 ) dt, for |x| < 1
ln(1t +
− tsin
t
dt, for all x
0
t
0
solution Substituting t 2 for t in the Maclaurin series for ln(1 + t) yields
55.
ln(1 + t 2 ) =
∞
(−1)n−1
n=1
∞
(t 2 )n
t 2n
=
.
(−1)n
n
n
n=1
Thus,
x
0
ln(1 + t 2 ) dt =
∞
n=1
(−1)n
x
∞
t 2n+1 x 2n+1
.
(−1)n
=
n(2n + 1) n(2n + 1)
0
n=1
∞
x
dt has Maclaurin series
(−1)n 2n x n ?
57. Which function
, for |x| < 1
4
n=0
0
1−t
solution We recognize that
∞
(−1)n 2n x n =
n=0
∞
(−2x)n
n=0
1 with x replaced by −2x. Therefore,
is the Maclaurin series for 1−x
∞
(−1)n 2n x n =
n=0
1
1
=
.
1 − (−2x)
1 + 2x
In Exercises
59–62,
use has
Theorem
2 to prove
Which
function
Maclaurin
seriesthat the f (x) is represented by its Maclaurin series for all x.
x x ∞
59. f (x) = sin 2 + cos 3 ,
(−1)k
(x − 3)k ?
k+1
solution All derivatives of f(x) consist
of
sin
or
cos
applied
to each of x/2 and x/3 and added together, so each
3
k=0
(n) summand is bounded by 1. Thus f (x) ≤ 2 for all n and x. By Theorem 2, f (x) is represented by its Taylor series for
everyFor
x. which values of x is the expansion valid?
May 23, 2011
706
C H A P T E R 10
INFINITE SERIES
61. f (x) = sinh −x
x
f (x) = e ,
solution By definition, sinh x = 21 (ex − e−x ), so if both ex and e−x are represented by their Taylor series centered
at c, then so is sinh x. But the previous exercise shows that e−x is so represented, and the text shows that ex is.
In Exercises 63–66, find100
the functions with the following Maclaurin series (refer to Table 1 on page 599).
f (x) = (1 + x)
x9
x 12
x6
63. 1 + x 3 +
+
+
+ ···
2!
3!
4!
solution We recognize
1 + x3 +
∞ 3n
∞
x
(x 3 )n
x9
x 12
x6
+
+
+ ··· =
=
2!
3!
4!
n!
n!
n=0
n=0
as the Maclaurin series for ex with x replaced by x 3 . Therefore,
1 + x3 +
3
x9
x 12
x6
+
+
+ · · · = ex .
2!
3!
4!
53 x 3
55 x 5
57 x 7
65. 1 −1 − 4x +
− 43 x 3 ++4·4·x·4 − 45 x 5 + · · ·
+ 42 x 2 −
3!
5!
7!
solution Note
53 x 3
55 x 5
57 x 7
53 x 3
55 x 5
57 x 7
1−
+
−
+ · · · = 1 − 5x + 5x −
+
−
+ ···
3!
5!
7!
3!
5!
7!
= 1 − 5x +
∞
(−1)n
n=0
(5x)2n+1
.
(2n + 1)!
The series is the Maclaurin series for sin x with x replaced by 5x, so
1−
55 x 5
57 x 7
53 x 3
+
−
+ · · · = 1 − 5x + sin(5x).
3!
5!
7!
In Exercises 6712
and 68,20let 28
x
x
x
+
−
+ ···
x4 −
3
5
7
f (x) =
1
(1 − x)(1 − 2x)
67. Find the Maclaurin series of f (x) using the identity
f (x) =
solution
1
2
−
1 − 2x
1−x
Substituting 2x for x in the Maclaurin series for
1
gives
1−x
∞
∞
1
=
2n x n
(2x)n =
1 − 2x
n=0
n=0
1
is valid for |x| < 1, the two series
which is valid for |2x| < 1, or |x| < 12 . Because the Maclaurin series for
1−x
together are valid for |x| < 12 . Thus, for |x| < 12 ,
∞
∞
2
1
1
=
−
=2
2n x n −
xn
(1 − 2x)(1 − x)
1 − 2x
1−x
n=0
=
∞
n=0
2n+1 x n −
∞
n=0
xn =
n=0
∞ 2n+1 − 1 x n .
n=0
69. When a voltage V is applied to a series circuit consisting of a resistor R and an inductor L, the current at time t is
Find the Taylor series for f (x) at c = 2. Hint:Rewrite
the identity of Exercise 67 as
V I (t) =
2 1 − e−Rt/L 1
R
−
f (x) =
−3 − 2(x − 2) −1 − (x − 2)
Vt
for small t.
Expand I (t) in a Maclaurin series. Show that I (t) ≈
L
May 23, 2011
S E C T I O N 10.7
solution
Taylor Series
707
t
Substituting − Rt
L for t in the Maclaurin series for e gives
n
∞ − Rt
∞
∞
(−1)n R n n
(−1)n R n n
L
−Rt/L
e
=
=
t =1+
t
n!
n!
L
n!
L
n=0
n=0
n=1
Thus,
⎛
1 − e−Rt/L = 1 − ⎝1 +
∞
(−1)n R n
n!
n=1
L
⎞
t n⎠ =
∞
(−1)n+1 Rt n
n=1
n!
L
,
and
∞
∞
Vt
V (−1)n+1 Rt n
V (−1)n+1 Rt n
=
.
+
I (t) =
R
n!
L
L
R
n!
L
n=1
n=2
If t is small, then we can approximate I (t) by the first (linear) term, and ignore terms with higher powers of t; then we
find
V (t) ≈
Vt
.
L
3 ) and use it to determine f (6) (0).
71. Find
thethe
Maclaurin
for 69
f (x)
cos(x
Use
result ofseries
Exercise
and=your
knowledge
of alternating series to show that
solution The Maclaurin series for cos x is
R
Vt
Vt
1−
t ≤ I (t) ≤
(for all t)
L
2L L
∞
x 2n
n
cos x =
(−1)
(2n)!
n=0
Substituting x 3 for x gives
cos(x 3 ) =
∞
n=0
(−1)n
x 6n
(2n)!
Now, the coefficient of x 6 in this series is
−
1
1
f (6) (0)
=− =
2!
2
6!
so
f (6) (0) = −
6!
= −360
2
20
73.
first
of the
the Maclaurin
Maclaurin series.
series for f (x) = ex . How does the result
(0) and f (8)to
(0)find
for the
f (x)
= three
tan−1terms
x using
Find Use
f (7)substitution
(k)
show that f (0) = 0 for 1 ≤ k ≤ 19?
solution
Substituting x 20 for x in the Maclaurin series for ex yields
ex
20
=
∞
∞ 20n
(x 20 )n
x
=
;
n!
n!
n=0
n=0
the first three terms in the series are then
1
1 + x 20 + x 40 .
2
(k)
Recall that the coefficient of x k in the Maclaurin series for f is f k!(0) . For 1 ≤ k ≤ 19, the coefficient of x k in the
20
Maclaurin series for f (x) = ex is zero; it therefore follows that
f (k) (0)
=0
k!
or f (k) (0) = 0
for 1 ≤ k ≤ 19.
Use the binomial series to find f (8) (0) for f (x) =
May 23, 2011
1 − x2.
708
C H A P T E R 10
INFINITE SERIES
75. Does the Maclaurin series for f (x) = (1 + x)3/4 converge to f (x) at x = 2? Give numerical evidence to support
your answer.
solution The Taylor series for f (x) = (1 + x)3/4 converges to f (x) for |x| < 1; because x = 2 is not contained on
this interval, the series does not converge to f (x) at x = 2. The graph below displays
SN =
N 3 4
n
n=0
2n
for 0 ≤ N ≤ 14. The divergent nature of the sequence of partial sums is clear.
SN
15
10
5
0
2
−5
4
6
8
10 12 14
N
−10
−15
−20
√
77.
Let f (x) = 1 + x.
Explain the steps required to verify that the Maclaurin series for f (x) = ex converges to f (x) for all x.
(a) Use a graphing calculator to compare the graph of f with the graphs of the first five Taylor polynomials for f . What
do they suggest about the interval of convergence of the Taylor series?
(b) Investigate numerically whether or not the Taylor expansion for f is valid for x = 1 and x = −1.
solution
(a) The five first terms of the Binomial series with a = 12 are
1 1 −1
1 1 −1
1 1 −1
1 −2
1 −2
1 −3
√
1
2 2
2
2
2
2
2
2
2
1+x =1+ x+
x2 +
x3 +
x4 + · · ·
2
2!
3!
4!
9
45
1
1
= 1 + x − x2 + x3 − x4 + · · ·
2
8
4
2
Therefore, the first five Taylor polynomials are
T0 (x) = 1;
1
T1 (x) = 1 + x;
2
1
T2 (x) = 1 + x −
2
1
T3 (x) = 1 + x −
2
1
T4 (x) = 1 + x −
2
1 2
x ;
8
1 2
x +
8
1 2
x +
8
1 3
x ;
8
1 3
5 4
x −
x .
8
128
The figure displays the graphs of these Taylor polynomials, along with the graph of the function f (x) =
is shown in red.
1.5
1
1.5
–1
0.5
0.5
1
The graphs suggest that the interval of convergence for the Taylor series is −1 < x < 1.
N 1 2
x n for x = 1 we find
(b) Using a computer algebra system to calculate SN =
n
n=0
S10 = 1.409931183,
May 23, 2011
S100 = 1.414073048,
S1000 = 1.414209104,
√
1 + x, which
S E C T I O N 10.7
which appears to be converging to
Taylor Series
709
N 1 √
2
· (−1)n , and find
2 as expected. At x = −1 we calculate SN =
n
n=0
S10 = 0.176197052,
S100 = 0.056348479,
S1000 = 0.017839011,
which appears to be converging to zero, though slowly.
79. Use Example 11 and the approximation sin x ≈ x to show that the period T of a pendulum released at an angle θ has
Use the first five terms of the Maclaurin series for the elliptic function E(k) to estimate the period T of a 1-meter
the following second-order approximation:
pendulum released at an angle θ = π4 (see Example 11).
θ2
L
T ≈ 2π
1+
g
16
solution The period T of a pendulum of length L released from an angle θ is
L
E(k),
T =4
g
where g ≈ 9.8 m/s2 is the acceleration due to gravity, E(k) is the elliptic function of the first kind and k = sin θ2 . From
Example 11, we know that
∞ π 1 · 3 · 5 · · · (2n − 1) 2 2n
E(k) =
k .
2
2 · 4 · 6 · · · (2n)
n=0
Using the approximation sin x ≈ x, we have
k = sin
θ
θ
≈ ;
2
2
moreover, using the first two terms of the series for E(k), we find
"
2 2 #
π
1
θ2
θ
π
E(k) ≈
1+
1+
.
=
2
2
2
2
16
Therefore,
L
T =4
E(k) ≈ 2π
g
L
θ2
1+
.
g
16
In Exercises 80–83, find the Maclaurin series of the function and use it to calculate the limit.
3
sin x − x + x6 x 2
cos x − 1 + 2
81. lim
x→0lim
x5 4
x→0
x
solution Using the Maclaurin series for sin x, we find
sin x =
∞
(−1)n
n=0
∞
x3
x5
x 2n+1
x 2n+1
=x−
+
+
.
(−1)n
(2n + 1)!
6
120
(2n + 1)!
n=3
Thus,
∞
sin x − x +
x3
x5
x 2n+1
=
(−1)n
+
6
120
(2n + 1)!
n=3
and
3
sin x − x + x6
x5
∞
=
1
x 2n−4
+
(−1)n
120
(2n + 1)!
n=3
Note that the radius of convergence for this series is infinite, and recall from the previous section that a convergent power
series is continuous within its radius of convergence. Thus to calculate the limit of this power series as x → 0 it suffices
to evaluate it at x = 0:
⎞
⎛
3
∞
2n−4
sin x − x + x6
x
1
n
⎠= 1 +0= 1
lim
+
= lim ⎝
(−1)
(2n + 1)!
120
120
x→0
x→0 120
x5
n=3
May 23, 2011
710
C H A P T E R 10
INFINITE SERIES
2)
cos x
sin(x−1
tan x−− x cos x − 16 x 3
83. lim
x→0lim x 4
x 25
x→0
x
solution We start with
sin x =
∞
(−1)n
n=0
x 2n+1
(2n + 1)!
cos x =
∞
(−1)n
n=0
x 2n
(2n)!
so that
∞
∞
sin(x 2 )
x 4n+2
x 4n−2
n
=
(−1)
=
(−1)n
4
4
(2n + 1)!
x
(2n + 1)!x
n=0
n=0
cos x
=
x2
∞
(−1)n
n=0
x 2n−2
(2n)!
Expanding the first few terms gives
∞
sin(x 2 )
1
x 4n−2
= 2 −
(−1)n
4
(2n + 1)!
x
x
n=1
∞
cos x
1
1 x 2n−2
= 2 − +
(−1)n
2
2
(2n)!
x
x
n=2
so that
∞
∞
n=1
n=2
4n−2
2n−2
sin(x 2 ) cos x
1 n x
nx
−
−
−
=
(−1)
(−1)
2
(2n + 1)!
(2n)!
x4
x2
Note that all terms under the summation signs have positive powers of x. Now, the radius of convergence of the series
for both sin and cos is infinite, so the radius of convergence of this series is infinite. Recall from the previous section that
a convergent power series is continuous within its radius of convergence. Thus to calculate the limit of this power series
as x → 0 it suffices to evaluate it at x = 0:
⎞
⎛
∞
∞
4n−2
2n−2
sin(x 2 ) cos x
x
x
1
⎠= 1 +0= 1
lim
−
− 2
(−1)n
(−1)n
= lim ⎝ −
4
(2n + 1)!
(2n)!
2
2
x→0
x→0 2
x
x
n=1
n=2
Further Insights and Challenges
1
t
85. LetIn
g(t)
− show2 that
. the Maclaurin expansion of f (x) = ln(1 + x) is valid for x = 1.
this=exercise
we
1 + t2
1+t
(a) Show that
1 for all x =π−1, 1
g(t) dt = − ln 2.
(a) Show that
N
4
2
0
(−1)N+1 x N+1
1
n n
2
3
4
5
=
(−1)
−xt − t 6 +
... x +
(b) Show that g(t) = 1 − t − t + t − t1 +
1+x
(c) Evaluate S = 1 − 12 − 13 + 14 − 15 − 16 − 17 +n=0
...
(b) Integrate from 0 to 1 to obtain
solution
1 N+1
N
(−1)n−1
x
dx
N+1
+
(−1)
ln
2
=
1
1
n
1
+
x
π
1
1
0
n=1 ln(t 2 + 1) = tan−1 1 − ln 2 =
− ln 2
g(t) dt = tan−1 t −
2
2
4
2
0
0
(c) Verify that the integral on the right tends to zero as N → ∞ by showing that it is smaller than 01 x N+1 dx.
1 :
(b) Start
with the
series for 1+t
(d) Prove
the Taylor
formula
(a)
1∞ 1 1
ln 2 1= 1=
− +(−1)−n t n + · · ·
2 3 4
1+t
n=0
and substitute t 2 for t to get
∞
1
=
(−1)n t 2n = 1 − t 2 + t 4 − t 6 + . . .
2
1+t
n=0
May 23, 2011
S E C T I O N 10.7
Taylor Series
711
so that
∞
t
=
(−1)n t 2n+1 = t − t 3 + t 5 − t 7 + . . .
1 + t2
n=0
Finally,
g(t) =
1
t
−
= 1 − t − t2 + t3 + t4 − t5 − t6 + t7 + . . .
1 + t2
1 + t2
(c) We have
1
1
1
1
1
g(t) dt = (1 − t − t 2 + t 3 + t 4 − t 5 − . . . ) dt = t − t 2 − t 3 + t 4 + t 5 − t 6 − · · · + C
2
3
4
5
6
The radius of convergence of the series for g(t) is 1, so the radius of convergence of this series is also 1. However, this
series converges at the right endpoint, t = 1, since
1
1 1
1 1
1−
−
−
+
−
− ...
2
3 4
5 6
is an alternating series with general term decreasing to zero. Thus by part (a),
1−
1 1 1 1 1
π
1
− + + − − · · · = − ln 2
2 3 4 5 6
4
2
In Exercises 86 and 87, we investigate the convergence of the binomial series
∞ a
Ta (x) =
n=0
n
xn
87. By Exercise 86, Ta (x) converges for |x| < 1, but we do not yet know whether Ta (x) = (1 + x)a .
Prove that Ta (x) has radius of convergence R = 1 if a is not a whole number. What is the radius of convergence
(a) Verify the identity
if a is a whole number?
a
a
a
a
=n
+ (n + 1)
n
n
n+1
(b) Use (a) to show that y = Ta (x) satisfies the differential equation (1 + x)y = ay with initial condition y(0) = 1.
Ta (x)
is zero.
(c) Prove that Ta (x) = (1 + x)a for |x| < 1 by showing that the derivative of the ratio
(1 + x)a
solution
(a)
a (a − 1) · · · (a − n + 1)
a (a − 1) · · · (a − n + 1) (a − n)
a
a
=n·
+ (n + 1)
+ (n + 1) ·
n
n+1
n
n!
(n + 1)!
a (a − 1) · · · (a − n + 1) a (a − 1) · · · (a − n + 1) (a − n)
+
n!
(n − 1)!
a (a − 1) · · · (a − n + 1) (n + (a − n))
a
=
=a·
n
n!
=
(b) Differentiating Ta (x) term-by-term yields
Ta (x) =
∞ a
x n−1 .
n
n
n=1
Thus,
(1 + x)Ta (x) =
∞ ∞ ∞
∞ a
a
a
a
x n−1 +
xn =
xn +
xn
n
n
(n + 1)
n
n
n
n+1
n
n=1
=
n=1
∞ $
(n + 1)
n=0
a
n+1
n=0
+n
a
n
%
xn = a
n=0
Moreover,
Ta (0) =
May 23, 2011
n=0
∞ a
0
= 1.
a
n
x n = aTa (x).
712
C H A P T E R 10
INFINITE SERIES
(c)
d
dx
Ta (x)
(1 + x)a
=
(1 + x)Ta (x) − aTa (x)
(1 + x)a Ta (x) − a(1 + x)a−1 Ta (x)
=
= 0.
(1 + x)2a
(1 + x)a+1
Thus,
Ta (x)
= C,
(1 + x)a
for some constant C. For x = 0,
Ta (0)
1
= = 1, so C = 1.
a
(1 + 0)
1
Finally, Ta (x) = (1 + x)a .
2 2
89. Assume that a < b and let π/2
L bethe arc2 length
(circumference) of the ellipse xa + yb = 1 shown in Figure 5.
2
that for
The function G(k) = 0
1 − k sin t dt is called an elliptic function of the second kind. Prove
2 2
There
|k|is<no1,explicit formula for L, but it is known that L = 4bG(k), with G(k) as in Exercise 88 and k = 1 − a /b .
Use the first three terms of the expansion of Exercise 88 to estimate L when a = 4 and b = 5.
∞ π
π 1 · 3 · · · (2n − 1) 2 k 2n
y
G(k) = −
2
2
2n − 1
b 2 · · · 4 · (2n)
n=1
a
FIGURE 5 The ellipse
x 2
a
x
+
y 2
b
= 1.
solution With a = 4 and b = 5,
k=
and the arc length of the ellipse
x 2
4
+
y 2
5
42
3
1− 2 = ,
5
5
= 1 is
⎛
⎞
2 3 2n
∞ π
π
1
·
3
·
·
·
−
1)
3
(2n
5
⎜
⎟
= 20 ⎝ −
L = 20G
⎠.
5
2
2
2 · 4 · · · (2n)
2n − 1
n=1
Using the first three terms in the series for G(k) gives
243
36,157π
1 · 3 2 (3/5)4
9
1 2 (3/5)2
+
= 10π 1 −
−
=
≈ 28.398.
·
·
L ≈ 10π − 10π
2
1
2·4
3
100 40,000
4000
91. Irrationality of e Prove that e is an irrational number using the following argument by contradiction. Suppose that
Use Exercise 88 to prove that if a < b and a/b is near 1 (a nearly circular ellipse), then
e = M/N, where M, N are nonzero integers.
(a) Show that M! e−1 is a whole number.
π
a2 L
≈
3b
+
x
B such that M! e−1 equals
(b) Use the power series for e at x = −1 to show that there
2 is an integer
b
Hint: Use the first two terms of the series
for G(k).
1
1
−
+ ···
B + (−1)M+1
M + 1 (M + 1)(M + 2)
(c) Use your knowledge of alternating series with decreasing terms to conclude that 0 < |M! e−1 − B| < 1 and observe
that this contradicts (a). Hence, e is not equal to M/N .
solution Suppose that e = M/N , where M, N are nonzero integers.
(a) With e = M/N,
M!e−1 = M!
which is a whole number.
May 23, 2011
N
= (M − 1)!N,
M
Chapter Review Exercises
713
(b) Substituting x = −1 into the Maclaurin series for ex and multiplying the resulting series by M! yields
1
1
(−1)k
−1
M!e = M! 1 − 1 + − + · · · +
+ ··· .
2! 3!
k!
For all k ≤ M,
M!
is a whole number, so
k!
1
1
(−1)k
M! 1 − 1 + − + · · · +
2! 3!
M!
is an integer. Denote this integer by B. Thus,
(−1)M+2
1
(−1)M+1
1
M! e−1 = B + M!
+
+ · · · = B + (−1)M+1
−
+ ··· .
(M + 1)!
(M + 2)!
M + 1 (M + 1)(M + 2)
(c) The series for M! e−1 obtained in part (b) is an alternating series with an = M!
n! . Using the error bound for an
alternating series and noting that B = SM , we have
M! e−1 − B ≤ aM+1 =
1
< 1.
M +1
This inequality implies that M! e−1 − B is not a whole number; however, B is a whole number so M! e−1 cannot be a
whole number. We get a contradiction to the result in part (a), which proves that the original assumption that e is a rational
number is false.
Use the result of Exercise 73 in Section 4.5 to show that the Maclaurin series of the function
2
e−1/x for x = 0
f (x) =
n−3
0
forinxeach
= 0 sequence.
1. Let an =
and bn = an+3 . Calculate the first three
terms
n!
whose
bn Maclaurin series converges but does not converge
(a) aisn2T (x) = 0. This provides an example of a function f (x)(b)
to f (x) (except at x = 0).
(c) an bn
(d) 2an+1 − 3an
CHAPTER REVIEW EXERCISES
solution
(a)
1−3 2
= (−2)2 = 4;
1!
2−3 2
1 2
1
=
= −
= ;
2!
2
4
3−3 2
=
= 0.
3!
a12 =
a22
a32
(b)
4−3
1
=
;
4!
24
1
5−3
=
;
b2 = a5 =
5!
60
1
6−3
=
.
b3 = a6 =
6!
240
b1 = a4 =
(c) Using the formula for an and the values in (b) we obtain:
1−3
1!
2−3
a2 b2 =
2!
3−3
a3 b3 =
3!
a1 b1 =
May 23, 2011
1
1
=− ;
24
12
1
1
·
=−
;
60
120
1
·
= 0.
240
·
714
C H A P T E R 10
INFINITE SERIES
(d)
1
− 3(−2) = 5;
2a2 − 3a1 = 2 −
2
3
1
= ;
2a3 − 3a2 = 2 · 0 − 3 −
2
2
2a4 − 3a3 = 2 ·
1
1
−3·0=
.
24
12
In Exercises 3–8, compute2n
the−limit
lim an = 2.
2 state that it does not exist) assuming that n→∞
1 (or
Prove that lim
= using the limit definition.
n→∞ 3n + 2
3
3. lim (5an − 2an2 )
n→∞
solution
2
5an − 2an2 = 5 lim an − 2 lim an2 = 5 lim an − 2 lim an = 5 · 2 − 2 · 22 = 2.
lim
n→∞
n→∞
n→∞
n→∞
n→∞
5. lim ean
n→∞
1
lim
n→∞
a
solution Thenfunction f (x) = ex is continuous, hence:
lim ean = elimn→∞ an = e2 .
n→∞
7. lim (−1)n an
n→∞
lim cos(πan )
n→∞
solution Because lim an = 0, it follows that lim (−1)n an does not exist.
n→∞
n→∞
In Exercises 9–22,
determine
the limit of the sequence or show that the sequence diverges.
an +
n
lim
√
√
2
n→∞
9. an = n a+n 5+−n n + 2
solution
First rewrite an as follows:
√
√
√
√
n+5− n+2
n+5+ n+2
(n + 5) − (n + 2)
3
= √
= √
.
an =
√
√
√
√
n+5+ n+2
n+5+ n+2
n+5+ n+2
Thus,
3
lim an = lim √
= 0.
√
n→∞ n + 5 + n + 2
n→∞
2
11. an = 21/n 3
3n − n
an =The function
x
solution
1 − 2n3 f (x) = 2 is continuous, so
2
2
lim an = lim 21/n = 2limn→∞ (1/n ) = 20 = 1.
n→∞
n→∞
m
13. bm = 1 + (−1)
10n
an =Because 1 + (−1)m is equal to 0 for m odd and is equal to 2 for m even, the sequence {bm } does not approach
solution
n!
one limit; hence this sequence diverges.
n+2
15. bn = tan−1
1 + (−1)m
n+5
bm =
m
solution The function tan−1 x is continuous, so
π
n+2
n+2
lim bn = lim tan−1
= tan−1 lim
= tan−1 1 = .
n→∞
n→∞
n→∞ n + 5
n+5
4
2
17. bn = n2100
+ nn − 3 n
+ π+n 1
−
an =
solution Rewrite
n! bn as5n
n2 + n − n2 + 1
n2 + n + n2 + 1
n2 + n − n2 + 1
n−1
.
bn =
= = 2
2
2
2
2
n + n + n2 + 1
n +n+ n +1
n +n+ n +1
May 23, 2011
Chapter Review Exercises
Then
lim bn = lim n→∞
n→∞
n − 1
1 − n1
1−0
1
n
n
= lim = √
= .
√
2
2
n→∞
2
n + n + n + 1
1+0+ 1+0
1 + n1 + 1 + 12
n2
n2
n2
n2
n
1 3m 2
19. bm c=n =1 +n + n − n2 − n
m
1 m
solution
lim bm = lim 1 +
= e.
m→∞
m→∞
m
21. bn = n ln(n
ln n
+ 1)−
3 n
1+
cn =Write
solution
n
bn = n ln
n+1
n
=
ln 1 + n1
1
n
.
Using L’Hôpital’s Rule, we find
−1 ln 1 + n1
ln 1 + x1
· − 12
1 + x1
1 −1
x
lim bn = lim
=
lim
=
lim
=
lim
= 1.
1
+
1
1
n→∞
n→∞
x→∞
x→∞
x→∞
x
− 12
n
x
x
23. Use the Squeeze
to show that lim
ln(n2 +Theorem
1)
n→∞
cn =
ln(n3 + 1)
solution For all x,
−
arctan(n2 )
= 0.
√
n
π
π
< arctan x < ,
2
2
so
π/2
arctan(n2 )
π/2
−√ <
< √ ,
√
n
n
n
for all n. Because
π/2
−√
n→∞
n
lim
π/2
= lim √ = 0,
n→∞ n
it follows by the Squeeze Theorem that
arctan(n2 )
= 0.
√
n→∞
n
lim
an+1
1 n 1 n
3 − {a
2 . } such that {sin an } is convergent.
an = sequence
25. Calculate
Give anlim
example of, where
a divergent
n→∞ an
2
3 n
solution Because
3n
1 n 1 n 1 n 1 n
3 − 2 ≥ 3 − 3 =
2
3
2
3
6
and
lim
3n
n→∞ 6
= ∞,
we conclude that limn→∞ an = ∞, so L’Hôpital’s rule may be used:
1 3n+1 −
a
lim n+1 = lim 2 1
n
n→∞ an
n→∞
23 −
n+1
1 2n+1
n+2 − 2n+2
3 − 2 23
3
3−0
3
= 3.
= lim n+1
= lim
n+1 =
1 2n
n+1
n→∞ 3
n→∞
1
−0
−2
3
1 − 23
∞
n−2
= sums
an + S64with
Definethe
an+1
27. Calculate
partial
and aS17 =
of 2.
the series
.
n2 + 2n
n=1
(a) Compute an for n = 2, 3, 4, 5.
solution
(b) Show that {an } is increasing and is bounded by 3.
1
2
11
1
+ 0 find
+ its+value.= −
= −0.183333;
− and
(c) Prove that lim Sa4n =
exists
n→∞
3
15 24
60
1
2
3
4
5
287
1
+
+
+
+
=
= 0.065079.
S7 = − + 0 +
3
15 24 35 48 63
4410
May 23, 2011
715
716
C H A P T E R 10
INFINITE SERIES
8
16
32
4
29. Find the sum + 1+ 1 + 1 + · · · .
9 1−
27 +81 −243 + · · · .
Find the sum
4 42
43
solution This is a geometric series with common ratio r = 23 . Therefore,
4
4
4
8
16
32
+
+
+
+ ··· = 9 2 = .
9 27 81 243
3
1− 3
∞
n+3
∞2 n
31. Find the sum
2.
3n
.
Find the sum
n=−1
e
n=2
solution Note
∞
∞ n
∞
2n+3
2n
2
3
=
2
=
8
;
n
n
3
3
3
n=−1
therefore,
n=−1
n=−1
∞
1
3
2n+3
=8· ·
= 36.
3n
2 1− 2
3
n=−1
∞
∞
∞
∞
divergent
an and π bn such that
(an + bn ) = 1.
33. Give an example
of
series
−1
2
b − tan n diverges if b = .
Show that
n=1
n=1
n=1
2
n=1 n
solution Let an = 12 + 1, bn = −1. The corresponding series diverge by the Divergence Test; however,
∞
(an + bn ) =
n=1
∞ n
1
n=1
2
=
1
2
= 1.
1 − 21
∞
1
∞ 35. Evaluate S = 1
1 .
n(n
+
3)
−
. Compute SN for N
Let S =
n=3 n
n+2
n=1
solution Note that
3
1
SN = =
n(n + 3) 2
= 1, 2, 3, 4. Find S by showing that
1 1
1 11
−
− −
3 Nn+ 1n +N3 + 2
so that
N
n=3
N 1 1
1
1
=
−
n(n + 3)
3
n n+3
n=3
1 1
1 1
1 1
−
+
−
+
−
3 6
4 7
5 8
1
1
1
1
1 1
−
+ ··· +
−
+
−
6 9
N −1 N +2
N
N +3
1
1
1
1 1 1 1
+ + −
−
−
=
3 3 4 5 N +1 N +2 N +3
=
1
3
Thus
∞
n=3
N 1
1
=
lim
n(n + 3)
3 N→∞
1
=
3
n=3
1
1
−
n n+3
1 1 1
1
1
1
+ + −
−
−
3 4 5 N +1 N +2 N +3
1
=
3
1 1 1
+ +
3 4 5
In Exercises
usearea
the of
Integral
Test to many
determine
whether
infinite
converges.
Find37–40,
the total
the infinitely
circles
on thethe
interval
[0,series
1] in Figure
1.
37.
∞
n2
n=1
n3 + 1
solution
2
Let f (x) = 3x . This function is continuous and positive for x ≥ 1. Because
x +1
f (x) =
May 23, 2011
(x 3 + 1)(2x) − x 2 (3x 2 )
x(2 − x 3 )
=
,
(x 3 + 1)2
(x 3 + 1)2
=
47
180
Chapter Review Exercises
717
we see that f (x) < 0 and f is decreasing on the interval x ≥ 2. Therefore, the Integral Test applies on the interval x ≥ 2.
Now,
R
∞
x2
x2
1
3 + 1) − ln 9 = ∞.
ln(R
lim
dx
=
lim
dx
=
3 R→∞
R→∞ 2 x 3 + 1
2 x3 + 1
The integral diverges; hence, the series
∞
n=2
n=1
∞
1
∞
n2
(n + 2)(ln(n
+ 2))3
n=1
(n3 + 1)1.01
n=1
1
solution Let f (x) =
3
39.
∞
n2
n2
diverges,
as
does
the
series
.
n3 + 1
n3 + 1
(x+2) ln (x+2)
∞
0
1 dx, we have
. Using the substitution u = ln(x + 2), so that du = x+2
∞
∞
1
1
1 R
f (x) dx =
du = lim
du = lim − 2 R→∞ ln 2 u3
R→∞
2u ln 2
ln 2 u3
1
1
1
−
=
= lim
R→∞ 2(ln 2)2
2(ln R)2
2(ln 2)2
Since the integral of f (x) converges, so does the series.
In Exercises
use the Comparison or Limit Comparison Test to determine whether the infinite series converges.
∞ 41–48,
n3
∞
1n4
n=1 e
41.
(n + 1)2
n=1
For all n ≥ 1,
solution
0<
The series
1
1
<
n+1
n
so
1
1
< 2.
2
(n + 1)
n
∞
∞
1
1
is
a
convergent
p-series,
so
the
series
converges by the Comparison Test.
n2
(n + 1)2
n=1
n=1
∞
2
∞n + 1
43.
1
n3.5√− 2
n=2
n+n
n=1
2
1 . Now,
solution Apply the Limit Comparison Test with an = n3.5+1 and bn = 1.5
n −2
n
n2 +1
3.5
n3.5 + n1.5
L = lim n 1−2 = lim
= 1.
n→∞
n→∞ n3.5 − 2
n1.5
Because L exists and
∞
1
is a convergent p-series, we conclude by the Limit Comparison Test that the series
n1.5
n=1
∞
n2 + 1
also converges.
n3.5 − 2
n=2
∞
∞ n
1
5
n=2 nn +
− 5ln n
n=1
solution For all n ≥ 2,
45.
The series
∞
n=2
47.
n
n
1
< 5/2 = 3/2 .
n
n
n5 + 5
1
is a convergent p-series, so the series
n3/2
∞
n
converges by the Comparison Test.
5
n=2 n + 5
∞
n10 + 10n
∞
1
n11 +
11n
n=1
3n − 2n
n
n=1
10
n
solution Apply the Limit Comparison Test with an = n11 +10n and bn = 10
11 . Then,
n +11
n10 +10n
n
n +10
n +1
11
n
an
n
n
= lim n +11
L = lim
= 1.
n = lim 1110 n = lim 1011
n→∞ bn
n→∞
n→∞
n→∞
n
+11
n
10
11n
11n + 1
11
May 23, 2011
10
10
718
C H A P T E R 10
INFINITE SERIES
The series
∞ n
10
11
n=1
is a convergent geometric series; because L exists, we may therefore conclude by the Limit
Comparison Test that the series
∞ 10
n + 10n
also converges.
n11 + 11n
n=1
∞
∞ 20
n
49. Determine
convergence
of
n the
+ 21
n=1
n21 + 20n
n=1
n +n
2
solution With an = 3n −2 , we have
n
2n + n
using the Limit Comparison Test with bn = 23 .
n
3 −2
n
1
n
n
n
n
1
+
n
an
2 +n 3
6 + n3
2
L = lim
· n = lim n
= lim n
=
lim
n = 1
n→∞ bn
n→∞ 3 − 2 2
n→∞ 6 − 2n+1
n→∞
1
1−2 3
Since L = 1, the two series either both converge or both diverge. Since
Limit Comparison Test tells us that
∞ n
2 +n
also converges.
3n − 2
∞ n
2
n=1
3
is a convergent geometric series, the
n=1
∞
∞
1 .
lim lnan
an diverges. Hint: Show that a1n ≥ 2n
51. Let an = 1 − 1 − n1 . Show that n = 0 and that
.
Determine the convergence of n→∞ n using the Limit Comparison Test with bn =
n
n=1
1.5
1.4
n=1
solution
√
√
n−1
1
n− n−1
n − (n − 1)
1
1− 1− =1−
=
= √ √
=
√
√
n
n
n
n( n + n − 1)
n + n2 − n
≥
The series
n+
1
√
n2
=
1
.
2n
∞
1
1 also diverges by the Comparison Test.
diverges, so the series ∞
1
−
1
−
n=2
n
2n
n=2
∞
n
∞
.
53. Let S =
1
2 + 1)2
1 − 1 − 2 converges.
Determine(nwhether
n=1
n
(a) Show that S converges. n=2
Use Eq. (4) in Exercise 83 of Section 10.3 with M = 99 to approximate S. What is the maximum size of the
(b)
error?
solution
(a) For n ≥ 1,
The series
n
1
n
< 2 2 = 3.
(n2 + 1)2
(n )
n
∞
∞
1
n
is
a
convergent
p-series,
so
the
series
also converges by the Comparison Test.
n3
(n2 + 1)2
n=1
n=1
(b) With an = 2 n 2 , f (x) = 2 x 2 and M = 99, Eq. (4) in Exercise 83 of Section 10.3 becomes
(n +1)
(x +1)
∞
∞
100
n
x
n
x
+
dx
≤
S
≤
+
dx,
2
2
2
2
2
2
2
(n + 1)
(n + 1)
100 (x + 1)
100 (x + 1)2
n=1
n=1
99
⎞
∞
n
x
100
+
dx ⎠ ≤
.
0≤S−⎝
(n2 + 1)2
(x 2 + 1)2
(1002 + 1)2
100
n=1
⎛
or
99
Now,
99
n=1
n
= 0.397066274; and
(n2 + 1)2
R
1
x
x
1
1
dx
=
lim
dx
=
lim
−
+
2 R→∞
R→∞ 100 (x 2 + 1)2
R 2 + 1 1002 + 1
100 (x 2 + 1)2
∞
=
May 23, 2011
1
= 0.000049995;
20002
Chapter Review Exercises
719
thus,
S ≈ 0.397066274 + 0.000049995 = 0.397116269.
The bound on the error in this approximation is
100
= 9.998 × 10−7 .
(1002 + 1)2
In Exercises 54–57, determine whether the series converges absolutely. If it does not, determine whether it converges
conditionally.
∞
n
∞ (−1) n
(−1)
1.1
n √
ln(n + 1)
3
n=1
n + 2n
55.
n=1
solution
Consider the corresponding positive series
∞
n=1
1
. Because
n1.1 ln(n + 1)
1
n1.1 ln(n + 1)
and
1
< 1.1
n
∞
∞
1
(−1)n
is
a
convergent
p-series,
we
can
conclude
by
the
Comparison
Test
that
also converges.
n1.1
n1.1 ln(n + 1)
n=1
∞
Thus,
n=1
n=1
(−1)n
converges absolutely.
1.1
n ln(n + 1)
∞
cos π4 +π 2πn ∞
cos√ 4 + πn
n√
n=1
n
n=1
√
π
solution cos 4 + 2πn = cos π4 = 22 , so
57.
√ ∞
∞
cos π4 + 2πn
2 1
=
√
√ .
2
n
n
n=1
n=1
π
This is a divergent p-series, so the series
∞
cos 4 + 2πn
diverges.
√
n
n=1
∞
(−1)k ∞
.
59. Catalan’s constant is defined by K =
(−1)n
Use a computer algebra system to approximate
(2k + 1)2
√ to within an error of at most 10−5 .
3
k=0
n + n
n=1
(a) How many terms of the series are needed to calculate K with an error of less than 10−6 ?
(b)
Carry out the calculation.
solution
Using the error bound for an alternating series, we have
|SN − K| ≤
1
1
=
.
(2(N + 1) + 1)2
(2N + 3)2
For accuracy to three decimal places, we must choose N so that
1
< 5 × 10−3
(2N + 3)2
or (2N + 3)2 > 2000.
Solving for N yields
N>
1 √
2000 − 3 ≈ 20.9.
2
Thus,
K≈
21
k=0
May 23, 2011
(−1)k
= 0.915707728.
(2k + 1)2
720
C H A P T E R 10
INFINITE SERIES
∞
∞
∞
∞
an be an absolutely convergent series. Determine
whether the
following series
are convergent or divergent:
Give an example of conditionally convergent series
an and
bn such that
(an + bn ) converges abson=1
n=1 ∞
n=1
n=1
∞
1
(a) lutely.
(−1)n an
an + 2
(b)
n
61. Let
(c)
n=1
∞
1
(d)
1 + an2
n=1
solution
n=1
Because
∞
an converges absolutely, we know that
n=1
(a) Because we know that
∞ n=1
1
an + 2
n
∞
an converges and the series
n=1
n=1
(c) Because
an converges and that
∞
|an | converges.
n=1
∞
1
is a convergent p-series, the sum of these two series,
n2
n=1
also converges.
∞ ∞
(−1)n an =
|an |
n=1
∞
∞
|an |
n
n=1
(b) We have,
Because
n=1
∞
|an | converges, it follows that
∞
∞
n=1
(−1)n an converges absolutely, which implies that
n=1
∞
(−1)n an converges.
n=1
an converges, limn→∞ an = 0. Therefore,
n=1
lim
1
n→∞ 1 + a 2
n
and the series
∞
1
1 + an2
n=1
=
1
= 1 = 0,
1 + 02
diverges by the Divergence Test.
∞
∞
|an |
|an | converges, so the series
also converges by the Comparison Test.
(d) |ann | ≤ |an | and the series
n
n=1
n=1
√
In Exercises 63–70, apply the Ratio Test to determine convergence
or divergence, or state that the Ratio Test is inconclusive.
Let {an } be a positive sequence such that lim n an = 12 . Determine whether the following series converge or
n→∞
∞
n5
diverge:
63.
n
∞
∞
5∞
√
n=1
2an
(b)
3n an
(c)
an
(a)
5
solutionn=1With an = n5n ,
n=1
n=1
an+1 (n + 1)5 5n
1 5
1
=
1
+
·
=
,
a 5
n
n5
5n+1
n
and
an+1 1
1
1
1 5
=
ρ = lim lim 1 +
= ·1= .
n→∞ an
5 n→∞
n
5
5
Because ρ < 1, the series converges by the Ratio Test.
∞
1
∞ √
n+1
n
n2 + n83
n=1
n
n=1
solution With an =
65.
1
,
n2n +n3
2
2
n + n3
n2n 1 + n2n
an+1 1 + n2n
1
n
n2
=
= ·
·
,
a (n + 1)2n+1 + (n + 1)3 =
2
2 n + 1 1 + (n+1)2
n
(n + 1)2n+1 1 + (n+1)
n+1
n+1
2
and
a
1
1
ρ = lim n+1 = · 1 · 1 = .
n→∞ an
2
2
Because ρ < 1, the series converges by the Ratio Test.
May 23, 2011
2
Chapter Review Exercises
67.
721
∞ n2
2 4
∞
n
n!
n=1
n!
n=1
n2
solution With an = 2n! ,
(n+1)2
an+1 n!
22n+1
= 2
a (n + 1)! · n2 = n + 1
n
2
a
ρ = lim n+1 = ∞.
n→∞ an
and
Because ρ > 1, the series diverges by the Ratio Test.
∞ ∞n n 1
ln n
2
n!
n=1
n3/2
n=4
n 1
,
solution With an = n2 n!
n+1
n
an+1 1
2
1 n
1 n+1 n
1
= n+1
·
1
+
n!
=
=
,
a 2
(n + 1)!
n
2
n
2
n
n
69.
and
a
1
ρ = lim n+1 = e.
n→∞ an
2
Because ρ = 2e > 1, the series diverges by the Ratio Test.
In Exercises
the Root Test to determine convergence or divergence, or state that the Root Test is inconclusive.
∞ 71–74,
n n apply
1
∞
n!
1 4
n=1
71.
4n
n=1
solution With an = 41n ,
L = lim
n→∞
√
n a = lim
n
n→∞
n
1
1
= .
4n
4
Because L < 1, the series converges by the Root Test.
n
∞ ∞ 3 n
73.
2
4n
n=1
n
n=1
n
3
solution With an = 4n
,
L = lim
√
n
n→∞
an = lim
n
n→∞
3 n
3
= lim
= 0.
n→∞
4n
4n
Because L < 1, the series converges by the Root Test.
In Exercises
determine
convergence or divergence using any method covered in the text.
3
∞ 75–92,
1 n
cos
∞
2 n n
n=1
75.
3
n=1
solution This is a geometric series with ratio r = 23 < 1; hence, the series converges.
77.
∞
∞−0.02n
e π 7n
n=1
e8n
n=1
solution This is a geometric series with common ratio r =
79.
1 ≈ 0.98 < 1; hence, the series converges.
e0.02
∞
n−1
∞ (−1)
√
√ −0.02n
nne
+ n+1
n=1
n=1
solution
In this alternating series, an = √
1√
. The sequence {an } is decreasing, and
n+ n+1
lim an = 0;
n→∞
therefore the series converges by the Leibniz Test.
May 23, 2011
722
C H A P T E R 10
INFINITE SERIES
81.
∞
n
(−1)
∞
1
ln n
n=2
n(ln n)3/2
n=10
solution The sequence an = ln1n is decreasing for n ≥ 10 and
lim an = 0;
n→∞
therefore, the series converges by the Leibniz Test.
83.
∞
∞ n1
√e
n n + ln n
n=1
n!
n=1
solution
For n ≥ 1,
1
1
1
≤ √ = 3/2 .
√
n n
n
n n + ln n
The series
∞
n=1
85.
1
is a convergent p-series, so the series
n3/2
∞ ∞ 1
1
√ −√
1
n3
√
√
n
+1
n=1
n(1 + n)
∞
1
converges by the Comparison Test.
√
n
n
+ ln n
n=1
n=1
solution This series telescopes:
∞ 1
1
1
1
1
1
1
+ √ −√
+ √ −√
+ ...
= 1− √
√ −√
n
n+1
2
2
3
3
4
n=1
so that the nth partial sum Sn is
1
1
1
1
1
1
1
1
Sn = 1 − √
+ √ −√
+ √ −√
+ ··· + √ − √
=1− √
n
n+1
n+1
2
2
3
3
4
and then
∞ 1
1
1
= lim Sn = 1 − lim √
=1
√ −√
n→∞
n→∞
n
n+1
n+1
n=1
87.
∞
∞ 1
√
n + ln nn − ln(n + 1)
n=1
n=1
solution
For n ≥ 1,
√
n ≤ n, so that
∞
n=1
∞
1
1
√ ≥
2n
n+ n
n=1
which diverges since it is a constant multiple of the harmonic series. Thus
∞
n=1
1
√ diverges as well, by the Comparison
n+ n
Test.
∞
∞1
cos(πn)
nln n 2/3
n=2
n
n=2
solution For n ≥ N large enough, ln n ≥ 2 so that
89.
∞
n=N
1
nln n
≤
n=N
which is a convergent p-series. Thus by the Comparison Test,
n < N does not affect convergence.
∞
1
ln3 n
n=2
May 23, 2011
∞
1
n2
∞
n=N
1
nln n
also converges; adding back in the terms for
Chapter Review Exercises
91.
∞
sin2
n=1
π
n
For all x > 0, sin x < x. Therefore, sin2 x < x 2 , and for x = πn ,
solution
sin2
The series
723
1
π
π2
< 2 = π2 · 2 .
n
n
n
∞
∞
1
π
is
a
convergent
p-series,
so
the
series
sin2 also converges by the Comparison Test.
2
n
n
n=1
n=1
In Exercises
find the interval of convergence of the power series.
∞ 93–98,
22n
∞ n n
2 x n!
n=0
93.
n!
n=0
n n
solution With an = 2 n!x ,
2n+1 x n+1
a
2
n! · n n = lim x · = 0
ρ = lim n+1 = lim n→∞ an
n→∞ (n + 1)!
2 x n→∞
n
Then ρ < 1 for all x, so that the radius of convergence is R = ∞, and the series converges for all x.
95.
∞
6
∞n n
x (x − 3)n
8
n +1
n=0
n+1
n=0
6
n
solution With an = n (x−3)
,
n8 +1
(n + 1)6 (x − 3)n+1
an+1 n8 + 1 ρ = lim = lim · 6
n→∞ an n→∞ (n + 1)8 − 1
n (x − 3)n (n + 1)6 (n8 + 1) = lim (x − 3) · 6
n→∞ n ((n + 1)8 + 1) n14 + terms of lower degree = lim (x − 3) · 14
= |x − 3|
n→∞ n + terms of lower degree Then ρ < 1 when |x − 3| < 1, so the radius of convergence is 1, and the series converges absolutely for |x − 3| < 1, or
∞
n6
, which converges by the Comparison Test comparing
2 < x < 4. For the endpoint x = 4, the series becomes
8
n +1
n=0
∞
∞ 6
1
n (−1)n
.
For
the
endpoint
x
=
2,
the
series
becomes
, which converges by the
with the convergent p-series
n2
n8 + 1
n=1
∞ 6
n (x − 3)n
therefore converges for 2 ≤ x ≤ 4.
Leibniz Test. The series
n8 + 1
n=0
n=0
97.
∞
∞
(nx)n n
nx
n=0
n=0
solution With an = nn x n , and assuming x = 0,
n (n + 1)n+1 x n+1 an+1 = lim x(n + 1) · n + 1 = ∞
=
lim
ρ = lim n
n
n→∞ an n→∞ n→∞
n x
n
n
1 n converges to e and the (n + 1) term diverges to ∞. Thus ρ < 1 only when x = 0, so the
=
1
+
since n+1
n
n
series converges only for x = 0.
2
∞ f (x) = n
99. Expand
(2x − 3)4 − 3x as a power series centered at c = 0. Determine the values of x for which the series converges.
n ln n
solution Write
n=0
1
1
2
=
.
4 − 3x
2 1 − 3x
4
May 23, 2011
724
C H A P T E R 10
INFINITE SERIES
1 , we obtain
Substituting 34 x for x in the Maclaurin series for 1−x
1
1 − 34 x
=
∞ n
3
n=0
4
xn.
This series converges for 34 x < 1, or |x| < 34 . Hence, for |x| < 43 ,
∞ 2
1 3 n n
=
x .
4 − 3x
2
4
n=0
∞
x 2k
.
101. LetProve
F (x) that
=
2k · k!
∞
k=0
(a) Show that F (x) has infinite radius of convergence.ne−nx =
n=0
(b) Show that y = F (x) is a solution of
e−x
(1 − e−x )2
Hint: Express the left-hand side as
the derivative of a geometric series.
y(0) = 1,
y (0) = 0
y = xy + y,
Plot the partial sums SN for N = 1, 3, 5, 7 on the same set of axes.
(c)
solution
2k
(a) With ak = xk ,
2 ·k!
ak+1 x2
|x|2k+2
2k · k!
a = 2k+1 · (k + 1)! · |x|2k = 2(k + 1) ,
k
and
ak+1 = x 2 · 0 = 0.
ρ = lim k→∞ ak Because ρ < 1 for all x, we conclude that the series converges for all x; that is, R = ∞.
(b) Let
y = F (x) =
∞
x 2k
.
2k · k!
k=0
Then
y =
∞
∞
2kx 2k−1
x 2k−1
=
,
2k k!
2k−1 (k − 1)!
k=1
k=1
∞
(2k − 1)x 2k−2
y =
,
2k−1 (k − 1)!
k=1
and
xy + y = x
∞
k=1
∞
∞
∞
x 2k−1
x 2k
x 2k
x 2k
=
+
+
1
+
2k k!
2k k!
2k−1 (k − 1)!
2k−1 (k − 1)!
k=0
k=1
k=1
∞
∞
∞
(2k + 1)x 2k
(2k + 1)x 2k
(2k − 1)x 2k−2
=
=
= y .
=1+
2k k!
2k k!
2k−1 (k − 1)!
k=1
k=0
k=1
Moreover,
y(0) = 1 +
∞
02k
=1
2k k!
k=1
Thus,
and
y (0) =
∞
k=1
02k−1
= 0.
2k−1 (k − 1)!
∞
x 2k
is the solution to the equation y = xy + y satisfying y(0) = 1, y (0) = 0.
2k k!
k=0
May 23, 2011
Chapter Review Exercises
(c) The partial sums S1 , S3 , S5 and S7 are plotted in the figure below.
y
7
6
5
4
3
2
1
−2
−1
1
2
x
In Exercises 103–112, find the Taylor series
centered at c.
∞
Find a power series P (x) =
an x n that satisfies the Laguerre differential equation
103. f (x) = e4x , c = 0
n=0
Substituting 4x for x in the Maclaurin series for ex yields
xy + (1 − x)y − y = 0
∞
∞ n
(4x)n
4 n
=
x .
with initial condition satisfying P (0) =e4x
1. =
n!
n!
solution
n=0
n=0
105. f (x) = x 4 , 2xc = 2
f (x) = e , c = −1
solution We have
f (x) = 4x 3
f (x) = 12x 2
f (x) = 24x
f (4) (x) = 24
and all higher derivatives are zero, so that
f (2) = 24 = 16 f (2) = 4 · 23 = 32 f (2) = 12 · 22 = 48 f (2) = 24 · 2 = 48
f (4) (2) = 24
Thus the Taylor series centered at c = 2 is
4
48
32
48
24
f (n) (2)
(x − 2)n = 16 + (x − 2) + (x − 2)2 + (x − 2)3 + (x − 2)4
n!
1!
2!
3!
4!
n=0
= 16 + 32(x − 2) + 24(x − 2)2 + 8(x − 2)3 + (x − 2)4
107. f (x) = sin x, c = π
f (x) = x 3 − x, c = −2
solution We have
f (4n) (x) = sin x
f (4n+1) (x) = cos x
f (4n+2) (x) = − sin x
f (4n+3) (x) = − cos x
so that
f (4n) (π) = sin π = 0 f (4n+1) (π) = cos π = −1
f (4n+2) (π) = − sin π = 0 f (4n+3) (π ) = − cos π = 1
Then the Taylor series centered at c = π is
∞
1
−1
−1
1
f (n) (π)
(x − π)n =
(x − π) + (x − π)3 +
(x − π)5 + (x − π )7 − . . .
n!
1!
3!
5!
7!
n=0
1
1
1
(x − π)5 +
(x − π )7 − . . .
= −(x − π) + (x − π)3 −
6
120
5040
1
109. f (x)
= = ex−1, , cc==−2
−1
f (x)
1 − 2x
solution Write
1
1
1
1
=
=
.
1 − 2x
5 − 2(x + 2)
5 1 − 2 (x + 2)
5
1 yields
Substituting 25 (x + 2) for x in the Maclaurin series for 1−x
1
1 − 25 (x + 2)
May 23, 2011
=
∞ n
2
(x + 2)n ;
5n
n=0
725
726
C H A P T E R 10
INFINITE SERIES
hence,
∞
∞
1
1 2n
2n
n=
=
(x
+
2)
(x + 2)n .
1 − 2x
5
5n
5n+1
n=0
x
111. f (x) = ln , c1 = 2
f (x) = 2
,
2
solution Write(1 − 2x)
n=0
c = −2
x
(x − 2) + 2
x−2
= ln
= ln 1 +
.
2
2
2
ln
Substituting x−2
2 for x in the Maclaurin series for ln(1 + x) yields
ln
n
∞ (−1)n+1 x−2
2
x
=
2
n=1
n
=
∞
(−1)n+1 (x − 2)n
.
n · 2n
n=1
This series is valid for |x − 2| < 2.
x first three terms of the Maclaurin series of f (x) and use it to calculate f (3) (0).
In Exercises 113–116,find the
, c=0
f (x) = x ln 1 +
2 2
2
x
113. f (x) = (x − x)e
solution
Substitute x 2 for x in the Maclaurin series for ex to get
2
1
1
ex = 1 + x 2 + x 4 + x 6 + . . .
2
6
so that the Maclaurin series for f (x) is
2
1
1
(x 2 − x)ex = x 2 + x 4 + x 6 + · · · − x − x 3 − x 5 − · · · = −x + x 2 − x 3 + x 4 + . . .
2
2
The coefficient of x 3 is
f (0)
= −1
3!
so that f (0) = −6.
1
115. f (x)
= = tan−1 (x 2 − x)
f (x)
1 + tan x
solution
1 to get
Substitute − tan x in the Maclaurin series for 1−x
1
= 1 − tan x + (tan x)2 − (tan x)3 + . . .
1 + tan x
We have not yet encountered the Maclaurin series for tan x. We need only the terms up through x 3 , so compute
tan (x) = sec2 x
tan (x) = 2(tan x) sec2 x
tan (x) = 2(1 + tan2 x) sec2 x + 4(tan2 x) sec2 x
so that
tan (0) = 1
tan (0) = 0
tan (0) = 2
Then the Maclaurin series for tan x is
tan x = tan 0 +
tan (0) 2 tan (0) 3
1
tan (0)
x+
x +
x + · · · = x + x3 + . . .
1!
2!
3!
3
Substitute these into the series above to get
2 3
1
1
1
1
= 1 − x + x3 + x + x3 − x + x3 + . . .
1 + tan x
3
3
3
1
= 1 − x − x 3 + x 2 − x 3 + higher degree terms
3
4
= 1 − x + x 2 − x 3 + higher degree terms
3
May 23, 2011
Chapter Review Exercises
The coefficient of x 3 is
4
f (0)
=−
3!
3
so that
f (0) = −6 ·
4
= −8
3
π
π7
π 3√
π5
117. Calculate
− x)3 1++ x5 − 7 + · · · .
f (x) = (sin
2
2 3! 2 5! 2 7!
solution We recognize that
∞
π
π7
π3
(π/2)2n+1
π5
− 3 + 5 − 7 + ··· =
(−1)n
2
(2n + 1)!
2 3! 2 5! 2 7!
n=0
is the Maclaurin series for sin x with x replaced by π/2. Therefore,
π7
π
π
π3
π5
− 3 + 5 − 7 + · · · = sin = 1.
2
2
2 3! 2 5! 2 7!
Find the Maclaurin series of the function F (x) =
May 23, 2011
x t
e −1
dt.
t
0
727
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